Structure of Atom
208 Questions
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th June Evening Shift
The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is :
[Given : The threshold frequency of platinum is 1.3 $\times$ 1015 s$-$1 and h = 6.6 $\times$ 10$-$34 J s.]
A.
3.21 $\times$ 10$-$14 J
B.
6.24 $\times$ 10$-$16 J
C.
8.58 $\times$ 10$-$19 J
D.
9.76 $\times$ 10$-$20 J
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th June Morning Shift
The pair, in which ions are isoelectronic with AI3+ is :
A.
Br$-$ and Be2+
B.
Cl$-$ and Li+
C.
S2$-$ and K+
D.
O2$-$ and Mg2+
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 24th June Evening Shift
The energy of one mole of photons of radiation of wavelength 300 nm is
(Given : h = 6.63 $\times$ 10$-$34 J s, NA = 6.02 $\times$ 1023 mol$-$1, c = 3 $\times$ 108 m s$-$1)
A.
235 kJ mol$-$1
B.
325 kJ mol$-$1
C.
399 kJ mol$-$1
D.
435 kJ mol$-$1
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 24th June Morning Shift
Consider the following pairs of electrons
(A) (a) n = 3, $l$ = 1, m1 = 1, ms = + ${1 \over 2}$
(b) n = 3, 1 = 2, m1 = 1, ms = + ${1 \over 2}$
(B) (a) n = 3, $l$ = 2, m1 = $-$2, ms = $-$${1 \over 2}$
(b) n = 3, $l$ = 2, m1 = $-$1, ms = $-$${1 \over 2}$
(C) (a) n = 4, $l$ = 2, m1 = 2, ms = + ${1 \over 2}$
(b) n = 3, $l$ = 2, m1 = 2, ms = + ${1 \over 2}$
The pairs of electrons present in degenerate orbitals is/are :
A.
Only (A)
B.
Only (B)
C.
Only (C)
D.
(B) and (C)
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 29th July Morning Shift
The minimum uncertainty in the speed of an electron in an one dimensional region of length $2 \mathrm{a}_{\mathrm{o}}$ (Where $\mathrm{a}_{\mathrm{o}}=$ Bohr radius $52.9 \,\mathrm{pm}$) is _________ $\mathrm{km} \,\mathrm{s}^{-1}$.
(Given : Mass of electron = 9.1 $\times$ 10$-$31 kg, Planck's constant h = 6.63 $\times$ 10$-$34 Js)
Correct Answer: 548
Explanation:
Heisenberg's uncertainty principle
$ \begin{aligned} & \Delta \mathrm{x} \times \Delta \mathrm{P}_{\mathrm{x}} \geq \frac{h}{4 \pi} \\ & \Rightarrow 2 \mathrm{a}_{0} \times \mathrm{m} \Delta \mathrm{v}_{\mathrm{x}}=\frac{h}{4 \pi} \text { (minimum) } \\ & \Rightarrow \Delta \mathrm{v}_{\mathrm{x}}=\frac{h}{4 \pi} \times \frac{1}{2 a_{0}} \times \frac{1}{m} \\ & =\quad 6.63 \times 10^{-34} \end{aligned} $
$ \begin{aligned} &4 \times 3.14 \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31} \\ & =548273 \,\mathrm{~ms}^{-1} \\ & =548.273 \,\mathrm{kms}^{-1} \\ & =548 \,\mathrm{kms}^{-1} \end{aligned} $
$ \begin{aligned} & \Delta \mathrm{x} \times \Delta \mathrm{P}_{\mathrm{x}} \geq \frac{h}{4 \pi} \\ & \Rightarrow 2 \mathrm{a}_{0} \times \mathrm{m} \Delta \mathrm{v}_{\mathrm{x}}=\frac{h}{4 \pi} \text { (minimum) } \\ & \Rightarrow \Delta \mathrm{v}_{\mathrm{x}}=\frac{h}{4 \pi} \times \frac{1}{2 a_{0}} \times \frac{1}{m} \\ & =\quad 6.63 \times 10^{-34} \end{aligned} $
$ \begin{aligned} &4 \times 3.14 \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31} \\ & =548273 \,\mathrm{~ms}^{-1} \\ & =548.273 \,\mathrm{kms}^{-1} \\ & =548 \,\mathrm{kms}^{-1} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th July Evening Shift
If the wavelength for an electron emitted from $\mathrm{H}$-atom is $3.3 \times 10^{-10} \mathrm{~m}$, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _________ times. (Nearest integer)
$\left[\right.$ Given $: \mathrm{h}=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ ]
Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$
Correct Answer: 2
Explanation:
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
$ \begin{aligned} &\Rightarrow \mathrm{mv}=\frac{\mathrm{h}}{\lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}} \times \mathrm{sec}}{3.3 \times 10^{-10} \mathrm{~m}} \\\\ &\mathrm{mv}=\frac{6.626 \times 10^{-24}}{3.3}=2 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \,\mathrm{sec}{ }^{-1} \end{aligned} $
Kinetic energy $=\frac{1}{2} m v^{2}$
$ \begin{aligned} &=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}} \\\\ &=\frac{\left(2 \times 10^{-24}\right)^{2}}{2 \times 9.1 \times 10^{-31} \mathrm{~kg}} \\\\ &=2.18 \times 10^{-18} \mathrm{~J} \\\\ &=21.8 \times 10^{-19} \mathrm{~J} \end{aligned} $
Total energy absorbed $=$ lonization energy $+$ Kinetic energy
$ \begin{aligned} &=(21.76+21.8) \times 10^{-19} \\\\ &=43.56 \times 10^{-19} \mathrm{~J} \\\\ &\approx 2 \text { times of } 21.76 \times 10^{-19} \mathrm{~J} \end{aligned} $
$ \begin{aligned} &\Rightarrow \mathrm{mv}=\frac{\mathrm{h}}{\lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}} \times \mathrm{sec}}{3.3 \times 10^{-10} \mathrm{~m}} \\\\ &\mathrm{mv}=\frac{6.626 \times 10^{-24}}{3.3}=2 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \,\mathrm{sec}{ }^{-1} \end{aligned} $
Kinetic energy $=\frac{1}{2} m v^{2}$
$ \begin{aligned} &=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}} \\\\ &=\frac{\left(2 \times 10^{-24}\right)^{2}}{2 \times 9.1 \times 10^{-31} \mathrm{~kg}} \\\\ &=2.18 \times 10^{-18} \mathrm{~J} \\\\ &=21.8 \times 10^{-19} \mathrm{~J} \end{aligned} $
Total energy absorbed $=$ lonization energy $+$ Kinetic energy
$ \begin{aligned} &=(21.76+21.8) \times 10^{-19} \\\\ &=43.56 \times 10^{-19} \mathrm{~J} \\\\ &\approx 2 \text { times of } 21.76 \times 10^{-19} \mathrm{~J} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th July Evening Shift
Consider an imaginary ion ${ }_{22}^{48} \mathrm{X}^{3-}$. The nucleus contains '$a$'% more neutrons than the number of electrons in the ion. The value of 'a' is _______________. [nearest integer]
Correct Answer: 4
Explanation:
Number of electrons in ${ }_{22}^{48} X^{3-}$ is 25 .
Number of neutrons $=48-22=26$.
$\%$ increase in the number of neutrons over electrons
$ =\left(\frac{26-25}{25}\right) 100=4 \% $
$\therefore a=4$
Number of neutrons $=48-22=26$.
$\%$ increase in the number of neutrons over electrons
$ =\left(\frac{26-25}{25}\right) 100=4 \% $
$\therefore a=4$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th July Morning Shift
The wavelength of an electron and a neutron will become equal when the velocity of the electron is $x$ times the velocity of neutron. The value of $x$ is ____________. (Nearest Integer)
(Mass of electron is $9.1 \times 10^{-31} \mathrm{~kg}$ and mass of neutron is $1.6 \times 10^{-27} \mathrm{~kg}$ )
Correct Answer: 1758
Explanation:
$\lambda_{e}=\frac{h}{m_{e} \times V_{e}}, \quad \lambda_{N}=\frac{h}{m_{N} \times V_{N}}$
$\lambda_{e}=\lambda_{N}$ When $V_{e}=x V_{N}$
$\frac{1}{m_{e} V_{e}}=\frac{1}{m_{N} \times V_{N}}$
$\frac{m_{N}}{m_{e}}=\frac{V_{e}}{V_{N}}=x$
$x=\frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}}$
$=0.17582 \times 10^{4}$
$\simeq 1758$
$\lambda_{e}=\lambda_{N}$ When $V_{e}=x V_{N}$
$\frac{1}{m_{e} V_{e}}=\frac{1}{m_{N} \times V_{N}}$
$\frac{m_{N}}{m_{e}}=\frac{V_{e}}{V_{N}}=x$
$x=\frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}}$
$=0.17582 \times 10^{4}$
$\simeq 1758$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th July Evening Shift
When the excited electron of a H atom from n = 5 drops to the ground state, the maximum number of emission lines observed are _____________.
Correct Answer: 10
Explanation:
Maximum number of emission lines
$ =\frac{\left(n_{2}-n_{1}\right)\left(n_{2}-n_{1}+1\right)}{2} $
$ \begin{aligned} &\mathrm{n}_{2}=5 \\ &\mathrm{n}_{1}=1 \\ &\Rightarrow \frac{(5-1)(5-1+1)}{2}=10 \end{aligned} $
Hence maximum number of emission lines observed are $10 .$
$ =\frac{\left(n_{2}-n_{1}\right)\left(n_{2}-n_{1}+1\right)}{2} $
$ \begin{aligned} &\mathrm{n}_{2}=5 \\ &\mathrm{n}_{1}=1 \\ &\Rightarrow \frac{(5-1)(5-1+1)}{2}=10 \end{aligned} $
Hence maximum number of emission lines observed are $10 .$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Morning Shift
If the work function of a metal is 6.63 $\times$ 10$-$19J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ____________ nm. (Nearest integer)
[Given : h = 6.63 $\times$ 10$-$34 J s, and c = 3 $\times$ 108 m s$-$1]
Correct Answer: 300
Explanation:
Given,
Work function = 6.63 $\times$ 10$-$19 J
$ = {{6.63 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$
= 4.14 eV
We know,
$E = {{1240} \over {\lambda \,(nm)}}$
$ \Rightarrow 4.14 = {{1240} \over \lambda }$
$\lambda$ = 300 nm
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Evening Shift
Consider the following set of quantum numbers.
| n | 1 | m$_1$ | |
|---|---|---|---|
| A. | 3 | 3 | $ - $3 |
| B. | 3 | 2 | $ - $2 |
| C. | 2 | 1 | +1 |
| D. | 2 | 2 | +2 |
The number of correct sets of quantum numbers is __________.
Correct Answer: 2
Explanation:
For A,
Given n = 3 and l = 3
but we know maximum value of l = n $-$ 1.
$\therefore$ l can't be equal to n.
So, Set A of quantum numbers is not possible.
For B,
Given n = 3, l = 2, m = $-$ 2
Here, l = 2 which follow the rule l = n $-$ 1.
And we know possible value of m is $-$ l to + l.
here possible value of m = $-$2 to +2
$\therefore$ This Set B is valid set of quantum numbers.
For C,
Given n = 2, l = 1, m = +1
Here l = 1 which follows the rule l = n $-$ 1.
For l = 1 possible value of m = $-$1 to +1
Here m = +1. So value of m is valid.
$\therefore$ Set C is valid set of quantum numbers.
For D,
Given n = 2, l = 2, m = +2
l = 2 does not follow the rule l = n $-$ 1 rule.
$\therefore$ Set D is not valid set of quantum numbers.
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Morning Shift
If the uncertainty in velocity and position of a minute particle in space are, 2.4 $\times$ 10$-$26 (m s$-$1) and 10$-$7 (m) respectively. The mass of the particle in g is ____________. (Nearest integer)
(Given : h = 6.626 $\times$ 10$-$34 Js)
Correct Answer: 22
Explanation:
We know from hisenberg uncertainty principle
$\Delta x\,.\,\Delta p = {h \over {4\pi }}$
$ \Rightarrow \Delta x\,.\,m\Delta v = {h \over {4\pi }}$
Given,
$\Delta$x = 10$-$7 m
$\Delta$x = 2.4 $\times$ 10$-$26 m/s
h = 6.626 $\times$ 10$-$34 Js
$\therefore$ ${10^{ - 7}} \times m \times 2.4 \times {10^{ - 26}} = {{6.626 \times {{10}^{ - 34}}} \over {4\pi }}$
$ \Rightarrow 2.4m = {{6.626} \over {4\pi \times 10}}$
$\Rightarrow$ m = 0.022 kg
$\Rightarrow$ m = 22 gm
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Morning Shift
The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is x $\times$ 10$-$8 m. The value of x is ___________. (Nearest Integer).
(Given : Energy of the electron in the first shell of the hydrogen atom is $-$2.2 $\times$ 10$-$18 J ; h = 6.63 $\times$ 10$-$34 Js and c = 3 $\times$ 108 ms$-$1)
Correct Answer: 4
Explanation:
Bohr model is not valid for lithium atom (Li) as Bohr model is valid for only single electronic species, so it would be valid for Li+2 but not Li atom.
So this question is BONUS.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
Given below are two statements.
Statement I : According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.
Statement II : According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.
Statement II : According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
Both Statement I and Statement II are false
B.
Both Statement I and Statement II are true
C.
Statement I is false but Statement II is true
D.
Statement I is true but Statement II is false
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Evening Shift
If the Thompson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been :
A.
All of the $\alpha$-particles pass through the gold foil without decrease in speed.
B.
$\alpha$-particles are deflected over a wide range of angles.
C.
All $\alpha$-particles get bounced back by 180$^\circ$
D.
$\alpha$-particles pass through the gold foil deflected by small angles and with reduced speed.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
Given below are two statements :
Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom.
Statement II : Bohr's model of hydrogen atom contradicts Heisenberg' uncertainty principle.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom.
Statement II : Bohr's model of hydrogen atom contradicts Heisenberg' uncertainty principle.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
Statement I is false but statement II is true.
B.
Statement I is true but statement II is false.
C.
Both statement I and statement II are false.
D.
Both statement I and statement II are true.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
Outermost electronic configuration of a group 13 element, E, is 4s2, 4p1. The electronic configuration of an element of p-block period-five placed diagonally to element, E is :
A.
[Kr] 3d10 4s2 4p2
B.
[Ar] 3d10 4s2 4p2
C.
[Xe] 5d10 6s2 6p2
D.
[Kr] 4d10 5s2 5p2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
Given below are two statements :
Statement I : Bohr's theory accounts for the stability and line spectrum of Li+ ion.
Statement II : Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Bohr's theory accounts for the stability and line spectrum of Li+ ion.
Statement II : Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
Statement I is false but statement II is true.
B.
Both statement I and statement II are true.
C.
Statement I is true but statement II is false.
D.
Both statement I and statement II are false.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
A certain orbital has no angular nodes and two radial nodes. The orbital is :
A.
2s
B.
3s
C.
3p
D.
2p
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
Which of the following forms of hydrogen emits low energy $\beta$- particles?
A.
Tritium $_1^3$H
B.
Proton H+
C.
Protium $_1^1$H
D.
Deuterium $_1^2$H
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
The orbital having two radial as well as two angular nodes is :
A.
3p
B.
5d
C.
4d
D.
4f
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
The plots of radial distribution functions for various orbitals of hydrogen atom against 'r' are given below :
The correct plot for 3s orbital is :
The correct plot for 3s orbital is :
A.
(B)
B.
(A)
C.
(C)
D.
(D)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
According to Bohr's atomic theory :
(A) Kinetic energy of electron is $ \propto {{{Z^2}} \over {{n^2}}}$.
(B) The product of velocity (v) of electron and principal quantum number (n), $'vn' \propto {Z^2}$.
(C) Frequency of revolution of electron in an orbit is $ \propto {{{Z^3}} \over {{n^3}}}$.
(D) Coulombic force of attraction on the electron is $ \propto {{{Z^3}} \over {{n^4}}}$.
Choose the most appropriate answer from the options given below :
(A) Kinetic energy of electron is $ \propto {{{Z^2}} \over {{n^2}}}$.
(B) The product of velocity (v) of electron and principal quantum number (n), $'vn' \propto {Z^2}$.
(C) Frequency of revolution of electron in an orbit is $ \propto {{{Z^3}} \over {{n^3}}}$.
(D) Coulombic force of attraction on the electron is $ \propto {{{Z^3}} \over {{n^4}}}$.
Choose the most appropriate answer from the options given below :
A.
(A), (C) and (D) only
B.
(A) only
C.
(C) only
D.
(A) and (D) only
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x $\times$ 1020. The value of x is __________. [Given : h = 6.63 $\times$ 10$-$34 Js and c = 3.0 $\times$ 108 ms$-$1]
Correct Answer: 2
Explanation:
Energy of photon is given as $E = {{nhc} \over \lambda }$ .... (i)
where, E = energy of photon (50 W),
n = number of photon
h = Planck's constant (6.63 $\times$ 10$-$34 Js)
c = speed of light (3 $\times$ 108 m/s)
$\lambda$ = wavelength of light (795 $\times$ 10$-$9 m)
E = 50W = 50 J = energy of photon
50 J = ${{n \times 6.63 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m/s} \over {795 \times {{10}^{ - 9}}m}}$
$\Rightarrow$ $n = {{50 \times 795 \times {{10}^{ - 9}}} \over {6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$
$ = 1998.49 \times {10^{17}} = 1.998 \times {10^{20}}$
$\Rightarrow$ $\approx$ 2 $\times$ 1020
$\therefore$ x = 2
where, E = energy of photon (50 W),
n = number of photon
h = Planck's constant (6.63 $\times$ 10$-$34 Js)
c = speed of light (3 $\times$ 108 m/s)
$\lambda$ = wavelength of light (795 $\times$ 10$-$9 m)
E = 50W = 50 J = energy of photon
50 J = ${{n \times 6.63 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m/s} \over {795 \times {{10}^{ - 9}}m}}$
$\Rightarrow$ $n = {{50 \times 795 \times {{10}^{ - 9}}} \over {6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$
$ = 1998.49 \times {10^{17}} = 1.998 \times {10^{20}}$
$\Rightarrow$ $\approx$ 2 $\times$ 1020
$\therefore$ x = 2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
Ge(Z = 32) in its ground state electronic configuration has x completely filled orbitals with ml = 0. The value of x is ___________.
Correct Answer: 7
Explanation:
Completely filled orbital with ml = 0 are
= 1 + 1 + 1 + 1 + 1 + 1 + 1
= 7
So, answer is 7.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x $\times$ 1013. The value of x is _____________. (Nearest integer)
(h = 6.63 $\times$ 10$-$34 Js, c = 3.00 $\times$ 108 ms$-$1)
(h = 6.63 $\times$ 10$-$34 Js, c = 3.00 $\times$ 108 ms$-$1)
Correct Answer: 50
Explanation:
Energy emitted in 0.1 sec.
= 0.1 sec. $\times$ ${10^{ - 3}}{J \over s}$
= 10$-$4 J
If 'n' photons of $\lambda$ = 1000 nm are emitted, then 10$-$4 = n $\times$ ${{hc} \over \lambda }$
$ \Rightarrow {10^{ - 4}} = {{n \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$
$\Rightarrow$ n = 5.02 $\times$ 1014 = 50.2 $\times$ 1013
$\Rightarrow$ 50 (nearest integer)
= 0.1 sec. $\times$ ${10^{ - 3}}{J \over s}$
= 10$-$4 J
If 'n' photons of $\lambda$ = 1000 nm are emitted, then 10$-$4 = n $\times$ ${{hc} \over \lambda }$
$ \Rightarrow {10^{ - 4}} = {{n \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$
$\Rightarrow$ n = 5.02 $\times$ 1014 = 50.2 $\times$ 1013
$\Rightarrow$ 50 (nearest integer)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to ${{{h^2}} \over {xma_0^2}}$. The value of 10x is ___________. (a0 is radius of Bohr's orbit) (Nearest integer) [Given : $\pi$ = 3.14]
Correct Answer: 3155
Explanation:
$mvr = {{nh} \over {2\pi }}$
$K.E. = {{{n^2}{h^2}} \over {8{\pi ^2}m{r^2}}} = {{4{h^2}} \over {8{\pi ^2}m{{(4{a_0})}^2}}}$
$ = \left( {{4 \over {8{\pi ^2} \times 16}}} \right){{{h^2}} \over {ma_0^2}}$
$\Rightarrow$ x = 315.507
$\Rightarrow$ 10x = 3155 (nearest integer)
$K.E. = {{{n^2}{h^2}} \over {8{\pi ^2}m{r^2}}} = {{4{h^2}} \over {8{\pi ^2}m{{(4{a_0})}^2}}}$
$ = \left( {{4 \over {8{\pi ^2} \times 16}}} \right){{{h^2}} \over {ma_0^2}}$
$\Rightarrow$ x = 315.507
$\Rightarrow$ 10x = 3155 (nearest integer)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3 $\times$ 1014 Hz. The velocity of ejected electron is ____________ $\times$ 105 ms$-$1 (Nearest integer)
[Use : h = 6.63 $\times$ 10$-$34 Js, me = 9.0 $\times$ 10$-$31 kg]
[Use : h = 6.63 $\times$ 10$-$34 Js, me = 9.0 $\times$ 10$-$31 kg]
Correct Answer: 5
Explanation:
$\upsilon $ : speed of electron having max. K.E.
$\Rightarrow$ from Einstein equation : E = $\phi$ + K.E.max
$ \Rightarrow {{hc} \over \lambda } = h{\upsilon _0} + {1 \over 2}m{v^2}$
$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {500 \times {{10}^{ - 9}}}} = 6.63 \times {10^{ - 34}} \times 4.3 \times {10^{14}} + {1 \over 2}m{v^2}$
$ \Rightarrow {{6.63 \times 30 \times {{10}^{ - 20}}} \over 5} = 6.63 \times 4.3 \times {10^{ - 20}} + {1 \over 2}m{v^2}$
$ \Rightarrow 11.271 \times {10^{ - 20}}J = {1 \over 2} \times 9 \times {10^{ - 31}} \times {\upsilon ^2}$
$\Rightarrow$ $\upsilon $ = 5 $\times$ 105 m/sec.
$\Rightarrow$ from Einstein equation : E = $\phi$ + K.E.max
$ \Rightarrow {{hc} \over \lambda } = h{\upsilon _0} + {1 \over 2}m{v^2}$
$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {500 \times {{10}^{ - 9}}}} = 6.63 \times {10^{ - 34}} \times 4.3 \times {10^{14}} + {1 \over 2}m{v^2}$
$ \Rightarrow {{6.63 \times 30 \times {{10}^{ - 20}}} \over 5} = 6.63 \times 4.3 \times {10^{ - 20}} + {1 \over 2}m{v^2}$
$ \Rightarrow 11.271 \times {10^{ - 20}}J = {1 \over 2} \times 9 \times {10^{ - 31}} \times {\upsilon ^2}$
$\Rightarrow$ $\upsilon $ = 5 $\times$ 105 m/sec.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
An accelerated electron has a speed of 5 $\times$ 106 ms$-$1 with an uncertainty of 0.02%. The uncertainty in finding its location while in motion is x $\times$ 10$-$9 m. The value of x is ____________. (Nearest integer)
[Use mass of electron = 9.1 $\times$ 10$-$31 kg, h =6.63 $\times$ 10$-$34 Js, $\pi$ = 3.14]
[Use mass of electron = 9.1 $\times$ 10$-$31 kg, h =6.63 $\times$ 10$-$34 Js, $\pi$ = 3.14]
Correct Answer: 58
Explanation:
$\Delta v = {{0.02} \over {100}} \times 5 \times {10^6} = {10^3}$ m/s
$\Delta x.\Delta v = {h \over {4\pi m}}$
$ \Rightarrow $ $x \times {10^{ - 9}} \times {10^3} = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}}}}$
$ \Rightarrow $ $x \times {10^{ - 9}} \times {10^3} = 0.058 \times {10^{ - 3}}$
$ \Rightarrow $ $x = {{0.058 \times {{10}^{ - 6}}} \over {{{10}^{ - 9}}}} = 58$
$\Delta x.\Delta v = {h \over {4\pi m}}$
$ \Rightarrow $ $x \times {10^{ - 9}} \times {10^3} = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}}}}$
$ \Rightarrow $ $x \times {10^{ - 9}} \times {10^3} = 0.058 \times {10^{ - 3}}$
$ \Rightarrow $ $x = {{0.058 \times {{10}^{ - 6}}} \over {{{10}^{ - 9}}}} = 58$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
A source of monochromatic radiation of wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, x $\times$ 1020 electrons are ejected per second. Assume that wavelength 400 nm is sufficient for ejection of electron from the surface of sodium metal. The value of x is ______________. (Nearest integer)
(h = 6.626 $\times$ 10$-$34 Js)
(h = 6.626 $\times$ 10$-$34 Js)
Correct Answer: 2
Explanation:
Total energy provided by
Source per second = ${{1000} \over {10}} = 100$ J
Energy required to eject electron = ${{hc} \over \lambda }$
= ${{6.626 \times {{10}^{ - 34}}} \over {400 \times {{10}^{ - 9}}}} \times 3 \times {10^8}$
Number of electrons ejected
= ${{100} \over {{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}}}$
= ${{400 \times {{10}^{ - 7}} \times {{10}^{26}}} \over {6.626 \times 3}}$
= ${{40 \times {{10}^{ - 20}}} \over {6.626 \times 3}}$
= $2.01 \times {10^{20}}$
Source per second = ${{1000} \over {10}} = 100$ J
Energy required to eject electron = ${{hc} \over \lambda }$
= ${{6.626 \times {{10}^{ - 34}}} \over {400 \times {{10}^{ - 9}}}} \times 3 \times {10^8}$
Number of electrons ejected
= ${{100} \over {{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}}}$
= ${{400 \times {{10}^{ - 7}} \times {{10}^{26}}} \over {6.626 \times 3}}$
= ${{40 \times {{10}^{ - 20}}} \over {6.626 \times 3}}$
= $2.01 \times {10^{20}}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
The wavelength of electrons accelerated from rest through a potential difference of 40 kV is x $\times$ 10$-$12 m. The value of x is ___________. (Nearest integer)
Give : Mass of electron = 9.1 $\times$ 10$-$31 kg
Charge on an electron = 1.6 $\times$ 10$-$19 C
Planck's constant = 6.63 $\times$ 10$-$34 Js
Give : Mass of electron = 9.1 $\times$ 10$-$31 kg
Charge on an electron = 1.6 $\times$ 10$-$19 C
Planck's constant = 6.63 $\times$ 10$-$34 Js
Correct Answer: 6
Explanation:
Wavelength of electron is given by
$\lambda = {h \over {\sqrt {2mqV} }}$
Here q = charge on electron, V = potential difference
$\lambda = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 40 \times {{10}^3}} }}$
$ = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {1164.8 \times {{10}^{ - 47}}} }} = 6.144 \times {10^{ - 12}} \approx 6 \times {10^{ - 12}}$
x = 6
OR
$\lambda = {{12.3} \over {\sqrt V }}\mathop A\limits^o $
$ = {{12.3} \over {200}} = 6.15 \times {10^{ - 12}}$ m
$ \therefore $ Ans. is 6
$\lambda = {h \over {\sqrt {2mqV} }}$
Here q = charge on electron, V = potential difference
$\lambda = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 40 \times {{10}^3}} }}$
$ = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {1164.8 \times {{10}^{ - 47}}} }} = 6.144 \times {10^{ - 12}} \approx 6 \times {10^{ - 12}}$
x = 6
OR
$\lambda = {{12.3} \over {\sqrt V }}\mathop A\limits^o $
$ = {{12.3} \over {200}} = 6.15 \times {10^{ - 12}}$ m
$ \therefore $ Ans. is 6
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
The Azimuthal quantum number for the valence electrons of Ga+ ion is ___________.
(Atomic number of Ga = 31)
(Atomic number of Ga = 31)
Correct Answer: 0
Explanation:
Ga = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1
Ga+ = 1s2 2s2 2p6 3s2 3p6 3d10 4s2
Azimuthal Quantum number (l) for valence shell electron is 0.
Ga+ = 1s2 2s2 2p6 3s2 3p6 3d10 4s2
Azimuthal Quantum number (l) for valence shell electron is 0.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
A certain orbital has n = 4 and mL = $-$3. The number of radial nodes in this orbital is ____________. (Round off to the Nearest Integer).
Correct Answer: 0
Explanation:
Number of radial nodes = n – $\ell $ – 1
n = 4, mL =–3 so $\ell $ = 3
radial nodes = 4 – 3 – 1 = 0
n = 4, mL =–3 so $\ell $ = 3
radial nodes = 4 – 3 – 1 = 0
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
The number of orbitals with n = 5, m1 = +2 is ___________. (Round off to the Nearest Integer).
Correct Answer: 3
Explanation:
Given, n = 5, ml = + 2
For n = 5, possible value of l = 0, 1, 2, 3, 4
For l = 0, ml = 0
l = 1, ml = $-$1, 0, 1
l = 2, ml = $-$2, $-$1, 0, 1, 2
l = 3, ml = $-$3, $-$2, $-$1, 0, 1, 2, 3
l = 4, ml = $-$4, $-$3, $-$2, $-$1, 0, 1, 2, 3, 4
Possible value of ml for a given value of l
= 0, $\pm$ 1, $\pm$ 2, $\pm$ 3 ..... $\pm$ l
So, number of orbitals having n = 5 and ml = $\pm$ 2 are 3.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _______$\mathop A\limits^o $. (Round off to the Nearest Integer).
[ Use : $\sqrt 3 $ = 1.73, h = 6.63 $\times$ 10$-$34 Js
me = 9.1 $\times$ 10$-$31 kg; c = 3.0 $\times$ 108 ms$-$1; 1eV = 1.6 $\times$ 10$-$19 J]
[ Use : $\sqrt 3 $ = 1.73, h = 6.63 $\times$ 10$-$34 Js
me = 9.1 $\times$ 10$-$31 kg; c = 3.0 $\times$ 108 ms$-$1; 1eV = 1.6 $\times$ 10$-$19 J]
Correct Answer: 9
Explanation:
$\lambda$ = 248 $\times$ 10$-$9 m
w0 = 3 $\times$ 1.6 $\times$ 10$-$19 J
E = w0 + K.E.
${{hc} \over \lambda } = {W_0} + K.E.$
$K.E. = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {248 \times {{10}^{ - 19}}}} - 3 \times 1.6 \times {10^{ - 19}}$
= 3.2 $\times$ 10$-$19 J
p = $\sqrt {2mK.E.} $
p = $\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 3.2 \times {{10}^{ - 19}}} $
p = 7.63 $\times$ 10$-$25
$ \therefore $ $\lambda = {h \over p} = {{6.626 \times {{10}^{ - 34}}} \over {7.63 \times {{10}^{ - 25}}}}$
$ \Rightarrow $ $\lambda$ = 8.7 $\times$ 10$-$10 = 8.7 $\mathop A\limits^o $ $ \approx $ 9 $\mathop A\limits^o $
w0 = 3 $\times$ 1.6 $\times$ 10$-$19 J
E = w0 + K.E.
${{hc} \over \lambda } = {W_0} + K.E.$
$K.E. = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {248 \times {{10}^{ - 19}}}} - 3 \times 1.6 \times {10^{ - 19}}$
= 3.2 $\times$ 10$-$19 J
p = $\sqrt {2mK.E.} $
p = $\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 3.2 \times {{10}^{ - 19}}} $
p = 7.63 $\times$ 10$-$25
$ \therefore $ $\lambda = {h \over p} = {{6.626 \times {{10}^{ - 34}}} \over {7.63 \times {{10}^{ - 25}}}}$
$ \Rightarrow $ $\lambda$ = 8.7 $\times$ 10$-$10 = 8.7 $\mathop A\limits^o $ $ \approx $ 9 $\mathop A\limits^o $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
A ball weighing 10 g is moving with a velocity of 90 ms$-$1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is ___________ $\times$ 10$-$33 m. (Rounded off to the nearest integer)
[Given : h = 6.63 $\times$ 10$-$34 Js]
[Given : h = 6.63 $\times$ 10$-$34 Js]
Correct Answer: 1
Explanation:
m = 10 g = 10$-$2 Kg
v = 90 m/sec.
$\Delta$ = v $\times$ 5% = 90 $\times$ ${5 \over {100}}$ = 4.5 m/sec
m . $\Delta$v . $\Delta$x $ \ge $ ${h \over {4\pi }}$
10$-$2 $\times$ 4.5 $\times$ $\Delta$x $ \ge $ ${{6.63 \times 3 \times {{10}^{ - 34}}} \over {4 \times {{22} \over 7}}}$
$\Delta$x $ \ge $ ${{6.63 \times 7 \times 2 \times {{10}^{ - 34}}} \over {9 \times 4 \times 22 \times {{10}^{ - 2}}}}$
$\Delta$x $ \ge $ 1.17 $\times$ 10$-$33 = x $\times$ 10$-$33
x = 1.17 $ \simeq $ 1
v = 90 m/sec.
$\Delta$ = v $\times$ 5% = 90 $\times$ ${5 \over {100}}$ = 4.5 m/sec
m . $\Delta$v . $\Delta$x $ \ge $ ${h \over {4\pi }}$
10$-$2 $\times$ 4.5 $\times$ $\Delta$x $ \ge $ ${{6.63 \times 3 \times {{10}^{ - 34}}} \over {4 \times {{22} \over 7}}}$
$\Delta$x $ \ge $ ${{6.63 \times 7 \times 2 \times {{10}^{ - 34}}} \over {9 \times 4 \times 22 \times {{10}^{ - 2}}}}$
$\Delta$x $ \ge $ 1.17 $\times$ 10$-$33 = x $\times$ 10$-$33
x = 1.17 $ \simeq $ 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization enegy of metal A in kJ mol$-$1 is __________. (Rounded off to the nearest integer)
[h = 6.63 $\times$ 10$-$34 Js, c = 3.00 $\times$ 108 ms$-$1, NA = 6.02 $\times$ 1023 mol$-$1]
[h = 6.63 $\times$ 10$-$34 Js, c = 3.00 $\times$ 108 ms$-$1, NA = 6.02 $\times$ 1023 mol$-$1]
Correct Answer: 181
Explanation:
Energy of EMR = IE of the metal (A)
$ = hv = {{hc} \over \lambda }$ atom$-$1 $ = {{hc} \over \lambda } \times {N_A}$ mol$-$1
$ = {{(6.63 \times {{10}^{ - 34}}) \times (3 \times {{10}^8}) \times (6.02 \times {{10}^{23}})} \over {(663 \times {{10}^{ - 9}})}}$ J mol$-$1 [$\because$ $\lambda$ = 663 nm = 663 $\times$ 10$-$9 m]
= 180600 J mol$-$1 = 180.6 kJ mol$-$1 $ \simeq $ 181 kJ mol$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
Among the following, number of metal/s which can be used as electrodes in the photoelectric cell is _________. (Integer answer)
(A) Li
(B) Na
(C) Rb
(D) Cs
(A) Li
(B) Na
(C) Rb
(D) Cs
Correct Answer: 1
Explanation:
Among the given alkali metals, only cesium (Cs)
is used as electrode in the photoelectric cell
due to its lowest ionisation energy.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
A proton and a Li3+ nucleus are accelerated by the same potential. If $\lambda _{Li}$ and $\lambda _p$ denote the
de Broglie wavelengths of Li3+ and proton respectively, then the value of
${{{\lambda _{Li}}} \over {{\lambda _p}}}$ is x $ \times $ 10-1.
The value of x is ______. (Rounded off to the nearest integer)
[Mass of Li3+ = 8.3 mass of proton]
The value of x is ______. (Rounded off to the nearest integer)
[Mass of Li3+ = 8.3 mass of proton]
Correct Answer: 2
Explanation:
Given, mass of Li3+ = 8.3 times of mass of proton formula,
De-Broglie wavelength, $\lambda = {h \over {\sqrt {2mqV} }}$
Here, h = Planck's constant = 6.624 $\times$ 10$-$34 J-s
m = Mass of atom
q = Charge (or number of electrons)
${\lambda _{Li}} = {h \over {\sqrt {2{m_{Li}}(3e)v} }}$ .... (i)
${\lambda _P} = {h \over {\sqrt {2{m_P}(e)v} }}$ .... (ii)
Now, Eq. (i) divided by Eq. (ii), we get,
${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P}(e)v} \over {{m_{Li}}(3e)v}}} $
We know that mLi = 8.3 mp
${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P} \times e \times V} \over {8.3\,{m_P} \times 3e \times V}}} = \sqrt {{1 \over {8.3 \times 3}}} = {1 \over 5} = 0.2$ = 2 $\times$ 10$-$1
Hence, x $\times$ 10$-$1, x = 2
De-Broglie wavelength, $\lambda = {h \over {\sqrt {2mqV} }}$
Here, h = Planck's constant = 6.624 $\times$ 10$-$34 J-s
m = Mass of atom
q = Charge (or number of electrons)
${\lambda _{Li}} = {h \over {\sqrt {2{m_{Li}}(3e)v} }}$ .... (i)
${\lambda _P} = {h \over {\sqrt {2{m_P}(e)v} }}$ .... (ii)
Now, Eq. (i) divided by Eq. (ii), we get,
${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P}(e)v} \over {{m_{Li}}(3e)v}}} $
We know that mLi = 8.3 mp
${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P} \times e \times V} \over {8.3\,{m_P} \times 3e \times V}}} = \sqrt {{1 \over {8.3 \times 3}}} = {1 \over 5} = 0.2$ = 2 $\times$ 10$-$1
Hence, x $\times$ 10$-$1, x = 2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
The correct statement about probability density (except at infinite distance from nucleus) is :
A.
It can be zero for 1s orbital
B.
It can be zero for 3p orbital
C.
It can never be zero for 2s orbital
D.
It can negative for 2p orbital
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
The difference between the radii of 3rd and 4th
orbits of Li2+ is R1 . The difference between the
radii of 3rd and 4th orbits of He+ is $\Delta $R2 .
Ratio $\Delta $R1 : $\Delta $R2 is :
orbits of Li2+ is R1 . The difference between the
radii of 3rd and 4th orbits of He+ is $\Delta $R2 .
Ratio $\Delta $R1 : $\Delta $R2 is :
A.
8 : 3
B.
3 : 2
C.
2 : 3
D.
3 : 8
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The shortest wavelength of H atom in the
Lyman series is $\lambda $1. The longest wavelength in
the Balmar series of He+ is :
A.
${{5{\lambda _1}} \over 9}$
B.
${{36{\lambda _1}} \over 5}$
C.
${{27{\lambda _1}} \over 5}$
D.
${{9{\lambda _1}} \over 5}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
The region in the electromagnetic spectrum
where the Balmar series lines appear is :
A.
Microwave
B.
Ultraviolet
C.
Visible
D.
Infrared
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
Consider the hypothetical situation where the
azimuthal quantum number,
$l$, takes values 0,
1, 2, ....., n + 1, where n is the principal
quantum number. Then, the element with
atomic number :
A.
13 has a half-filled valence subshell
B.
9 is the first alkali metal
C.
8 is the first noble gas
D.
6 has a 2p-valence subshell
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
The number of subshells associated with n = 4
and m = –2 quantum numbers is
A.
8
B.
2
C.
16
D.
4
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
The figure that is not a direct manifestation of
the quantum nature of atoms is :
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
The de Broglie wavelength of an electron in the
4th Bohr orbit is :
A.
2$\pi $a0
B.
6$\pi $a0
C.
8$\pi $a0
D.
4$\pi $a0
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
The radius of the second Bohr orbit, in terms
of the Bohr radius, a0, in Li2+ is :
A.
${{2{a_0}} \over 9}$
B.
${{2{a_0}} \over 3}$
C.
${{4{a_0}} \over 9}$
D.
${{4{a_0}} \over 3}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
Hydrogen has three isotopes (A), (B) and (C).
If the number of neutron(s) in (A), (B) and (C)
respectively, are (x), (y) and (z), the sum of (x),
(y) an (z) is :
A.
3
B.
1
C.
4
D.
2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
For the Balmer series in the spectrum of H atom,
$\overline \nu = {R_H}\left\{ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right\}$, the correct statements among (I) to (IV) are :
(I) As wavelength decreases, the lines in the series converge
(II) The integer n1 is equal to 2
(III) The lines of longest wavelength corresponds to n2 = 3
(IV) The ionization energy of hydrogen can be calculated from wave number of these lines
$\overline \nu = {R_H}\left\{ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right\}$, the correct statements among (I) to (IV) are :
(I) As wavelength decreases, the lines in the series converge
(II) The integer n1 is equal to 2
(III) The lines of longest wavelength corresponds to n2 = 3
(IV) The ionization energy of hydrogen can be calculated from wave number of these lines
A.
(II), (III), (IV)
B.
(I), (II), (III)
C.
(I), (III), (IV)
D.
(I), (II), (IV)