Structure of Atom
Assume that the radius of the first Bohr orbit of hydrogen atom is 0.6 $\mathrm{\mathop A\limits^o }$. The radius of the third Bohr orbit of He$^+$ is __________ picometer. (Nearest Integer)
Explanation:
Radius of 3rd Bohr orbit of He$^{+3}$
$ = 0.6 \times {{{{(3)}^2}} \over 2}$
$ = 0.3 \times 9$
$ = 2.7\,\mathop A\limits^o $
$ = 270 \times {10^{ - 12}}$ ppm
The number of given orbitals which have electron density along the axis is _________
$\mathrm{p_x,p_y,p_z,d_{xy},d_{yz},d_{xz},d_{z^2},d_{x^2-y^2}}$
Explanation:
Therefore, the total number of given orbitals which have electron density along the axis is 5.
If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is _________ nm. (Nearest integer)
Explanation:
$ \begin{aligned} & \Rightarrow R=\frac{9 \times 16}{720 \times 7} \\\\ & \frac{1}{\lambda^{\prime}}=\frac{9 \times 16}{720 \times 7} \times\left(\frac{1}{9}-\frac{1}{25}\right) \\\\ & \lambda^{\prime}=492.18 \mathrm{~nm} \\\\ & \lambda^{\prime}=492 \mathrm{~nm} \text { (nearest integer) } \end{aligned} $
[Use :
Bohr radius, $\mathrm{a}=52.9 \mathrm{pm}$
Rydberg constant, $R_{\mathrm{H}}=2.2 \times 10^{-18} \mathrm{~J}$
Planck's constant, $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
Speed of light, $\mathrm{c}=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$ ]
Explanation:
$r = \frac{52.9 \times n^2}{Z}$
where $r$ is in pm, $n$ is the principal quantum number, and $Z$ is the atomic number.
2. For a $\mathrm{He}^{+}$ ion, $Z=2$.
3. We are given the initial radius $r_2 = 105.8$ pm and the final radius $r_1 = 26.45$ pm.
4. We can substitute these radii into the equation for $r$ to find the corresponding quantum numbers.
For $r_2 = 105.8$ pm, we get :
$105.8 = \frac{52.9 \times n_2^2}{2}$
which gives $n_2 = 2$.
Similarly, for $r_1 = 26.45$ pm, we get :
$26.45 = \frac{52.9 \times n_1^2}{2}$
which gives $n_1 = 1$.
5. Therefore, the transition is from $n_2 = 2$ to $n_1 = 1$.
6. The energy difference during this transition is equal to the energy of the emitted photon, which is given by :
$E = \frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
where $h$ is the Planck's constant, $c$ is the speed of light, $R_H$ is the Rydberg constant, and $\lambda$ is the wavelength of the photon.
7. Substituting all known values into this equation gives :
$\frac{6.6 \times 10^{-34} \, \mathrm{J} \, \mathrm{s} \times 3 \times 10^8 \, \mathrm{m/s}}{\lambda} = 2.2 \times 10^{-18} \, \mathrm{J} \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$
Solving this equation yields $\lambda = 30 \times 10^{-9}$ m, or 30 nm.
8. So, the wavelength of the emitted photon during the transition is 30 nm.
Given below are the quantum numbers for 4 electrons.
A. $\mathrm{n}=3,l=2, \mathrm{~m}_{1}=1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
B. $\mathrm{n}=4,l=1, \mathrm{~m}_{1}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
C. $\mathrm{n}=4,l=2, \mathrm{~m}_{1}=-2, \mathrm{~m}_{\mathrm{s}}=-1 / 2$
D. $\mathrm{n}=3,l=1, \mathrm{~m}_{1}=-1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
The correct order of increasing energy is :
Given below are two statements: One is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$
Assertion $\mathbf{A}$ : Zero orbital overlap is an out of phase overlap.
Reason $\mathbf{R}$ : It results due to different orientation / direction of approach of orbitals.
In the light of the above statements, choose the correct answer from the options given below
Identify the incorrect statement from the following.
The correct decreasing order of energy for the orbitals having, following set of quantum numbers :
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1
is :
Outermost electronic configurations of four elements A, B, C, D are given below :
(A) $3 s^{2}$
(B) $3 s^{2} 3 p^{1}$
(C) $3 s^{2} 3 p^{3}$
(D) $3 s^{2} 3 p^{4}$
The correct order of first ionization enthalpy for them is :
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Energy of $2 \mathrm{s}$ orbital of hydrogen atom is greater than that of $2 \mathrm{s}$ orbital of lithium.
Reason R: Energies of the orbitals in the same subshell decrease with increase in the atomic number.
In the light of the above statements, choose the correct answer from the options given below.
Which of the following sets of quantum numbers is not allowed?
The number of radial nodes and total number of nodes in 4p orbital respectively are :
Which of the following is the correct plot for the probability density ${\psi ^2}$ (r) as a function of distance 'r' of the electron from the nucleus for 2s orbital?
Which of the following statements are correct?
(A) The electronic configuration of Cr is [Ar] 3d5 4s1.
(B) The magnetic quantum number may have a negative value.
(C) In the ground state of an atom, the orbitals are filled in order of their increasing energies.
(D) The total number of nodes are given by n $-$ 2.
Choose the most appropriate answer from the options given below :
Consider the following statements :
(A) The principal quantum number 'n' is a positive integer with values of 'n' = 1, 2, 3, ...
(B) The azimuthal quantum number 'l' for a given 'n' (principal quantum number) can have values as 'l' = 0, 1, 2, ...... n
(C) Magnetic orbital quantum number 'ml' for a particular 'l' (azimuthal quantum number) has (2l + 1) values.
(D) $\pm$ 1/2 are the two possible orientations of electron spin.
(E) For l = 5, there will be a total of 9 orbital
Which of the above statements are correct?
The number of radial and angular nodes in 4d orbital are, respectively
If the radius of the 3rd Bohr's orbit of hydrogen atom is r3 and the radius of 4th Bohr's orbit is r4. Then :
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.]
The pair, in which ions are isoelectronic with AI3+ is :
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)
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 :
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)
Explanation:
$ \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} $
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 $=9.1 \times 10^{-31} \mathrm{~kg}$
Explanation:
$ \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} $
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]
Explanation:
Number of neutrons $=48-22=26$.
$\%$ increase in the number of neutrons over electrons
$ =\left(\frac{26-25}{25}\right) 100=4 \% $
$\therefore a=4$
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 $1.6 \times 10^{-27} \mathrm{~kg}$ )
Explanation:
$\lambda_{e}=\lambda_{N}$ When $V_{e}=x V_{N}$
$\frac{1}{m_{e} V_{e}}=\frac{1}{m_{N} \times V_{N}}$
$\frac{m_{N}}{m_{e}}=\frac{V_{e}}{V_{N}}=x$
$x=\frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}}$
$=0.17582 \times 10^{4}$
$\simeq 1758$
When the excited electron of a H atom from n = 5 drops to the ground state, the maximum number of emission lines observed are _____________.
Explanation:
$ =\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 $10 .$
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]
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
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 __________.
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.
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)
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
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)
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.
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 : 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 : 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 :
The correct plot for 3s orbital is :
(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 :
Explanation:
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
Explanation:
Completely filled orbital with ml = 0 are
= 1 + 1 + 1 + 1 + 1 + 1 + 1
= 7
So, answer is 7.
(h = 6.63 $\times$ 10$-$34 Js, c = 3.00 $\times$ 108 ms$-$1)
Explanation:
= 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)
Explanation:
$K.E. = {{{n^2}{h^2}} \over {8{\pi ^2}m{r^2}}} = {{4{h^2}} \over {8{\pi ^2}m{{(4{a_0})}^2}}}$
$ = \left( {{4 \over {8{\pi ^2} \times 16}}} \right){{{h^2}} \over {ma_0^2}}$
$\Rightarrow$ x = 315.507
$\Rightarrow$ 10x = 3155 (nearest integer)
[Use : h = 6.63 $\times$ 10$-$34 Js, me = 9.0 $\times$ 10$-$31 kg]
Explanation:
$\Rightarrow$ 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.
[Use mass of electron = 9.1 $\times$ 10$-$31 kg, h =6.63 $\times$ 10$-$34 Js, $\pi$ = 3.14]
Explanation:
$\Delta x.\Delta v = {h \over {4\pi m}}$
$ \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 $ $x = {{0.058 \times {{10}^{ - 6}}} \over {{{10}^{ - 9}}}} = 58$
(h = 6.626 $\times$ 10$-$34 Js)
Explanation:
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
= ${{100} \over {{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}}}$
= ${{400 \times {{10}^{ - 7}} \times {{10}^{26}}} \over {6.626 \times 3}}$
= ${{40 \times {{10}^{ - 20}}} \over {6.626 \times 3}}$
= $2.01 \times {10^{20}}$
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
Explanation:
$\lambda = {h \over {\sqrt {2mqV} }}$
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
(Atomic number of Ga = 31)
Explanation:
Ga+ = 1s2 2s2 2p6 3s2 3p6 3d10 4s2
Azimuthal Quantum number (l) for valence shell electron is 0.