iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th April Evening Slot
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and
subsequently to the first excited state. The ratio of the respective wavelengths, ${{{\lambda _1}} \over {{\lambda _2}}}$, of the photons emitted
in this process is :
A.
${{22} \over 5}$
B.
${7 \over 5}$
C.
${9 \over 7}$
D.
${{20} \over 7}$
Correct Answer: D
Explanation:
n = 1 (Ground state)
n = 2 (First excitate state)
n = 3 (Second excitate state)
n = 4 (Third excitate state)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th April Evening Slot
Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 $\mathop A\limits^o $). The
de-Broglie wavelength of this electron is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th April Morning Slot
An excited He+
ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a
transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of
wavelength $\lambda $, energy $E = {{1240\,eV} \over {\lambda (in\,nm)}}$) :
A.
n = 4
B.
n = 7
C.
n = 5
D.
n = 6
Correct Answer: C
Explanation:
$\Delta {E_n} = - {{{E_0}{Z^2}} \over {{n^2}}}$
Let it start from n to m and from m to ground.
Then $13.6 \times 4\left| {1 - {1 \over {{m^2}}}} \right| = {{hc} \over {30.4\,nm}}$
$ \therefore $ m = 2, and now $13.6 \times 4\left( {{1 \over 4} - {1 \over {{n^2}}}} \right) = {{hc} \over {108.5 \times {{10}^{ - 9}}}}$
n = 5
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th April Evening Slot
In Li+ +, electron in first Bohr orbit is excited to a level by a radiation of wavelength $\lambda $. When the ion gets
deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral
lines are observed. What is the value of $\lambda $?
(Given : H = 6.63 × 10–34 Js; c = 3 × 108
ms
–1)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th April Evening Slot
Two radioactive substances A and B have decay constants 5$\lambda $ and $\lambda $ respectively. At t = 0, a sample has the
same number of the two nuclei. The time taken for the ratio of the number of nuclei to become ${\left( {{1 \over e}} \right)^2}$
will be :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th April Morning Slot
Two radioactive materials A and B have decay
constants 10$\lambda $ and $\lambda $, respectively. It initially
they have the same number of nuclei, then the
ratio of the number of nuclei of A to that of B
will be 1/e after a time :
$ \Rightarrow $ Ionisation energy = –T.E.
= 13.6 eV
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th April Morning Slot
Taking the wavelength of first Balmer line in
hydrogen spectrum (n = 3 to n = 2) as 660 nm,
the wavelength of the 2nd Balmer line (n = 4 to
n = 2) will be :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Evening Slot
The ratio of mass densities of nuclei of 40Ca
and 16O is close to :-
A.
1
B.
5
C.
0.1
D.
2
Correct Answer: A
Explanation:
Densities of nucleus happens to be constant,
irrespective of mass number.
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Morning Slot
Radiation coming from transitions
n = 2 to n = 1 of hydrogen atoms fall on He+
ions in n = 1 and n = 2 states. The possible
transition of helium ions as they absorb energy
from the radiation is :
A.
n = 1 $ \to $ n = 4
B.
n = 2 $ \to $ n = 5
C.
n = 2 $ \to $ n = 4
D.
n = 2 $ \to $ n = 3
Correct Answer: C
Explanation:
Energy released for tension n = 2 to n = 1 of hydrogen atom
(A) n = 1 to n = 4
$E = 13.6 \times {2^2} \times \left( {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{15} \over 4}eV$
(B) n = 2 to n = 4
$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{3} \over 4}eV$
(C) n = 2 to n = 5
$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{5^2}}}} \right) = 13.6 \times {{21} \over 25}eV$
(D) n = 2 to n = 3
$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = 13.6 \times {{5} \over 9}eV$
So, possible transition is n = 2 $\to$ n = 4
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th January Evening Slot
In a radioactive decay chain, the initial nucleus is ${}_{90}^{232}$Th. At the end there are 6 $\alpha $-particles and 4 $\beta $-particles which are emitted. If the end nucleus is ${}_Z^A$X, A and Z are given by :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th January Morning Slot
A particle of mass m moves in a circular orbit in a central potential field U(r) = ${1 \over 2}$ kr2. If Bohr 's
quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as :
A.
rn $ \propto $ $\sqrt n $, En $ \propto $ n
B.
rn $ \propto $ $\sqrt n $, En $ \propto $ ${1 \over n}$
$ \Rightarrow $ En = $ - {1 \over 2}U\left( r \right)$ = $ - {1 \over 4}k{r^2}$
$ \Rightarrow $ En $ \propto $ n
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 11th January Evening Slot
In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda $. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 11th January Morning Slot
A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980$\mathop A\limits^ \circ $. The radius of the atom in the excited state, in terms of Bohr radius a0 will be : (hc = 12500 eV$\mathop A\limits^ \circ $)
A.
4a0
B.
9a0
C.
25a0
D.
16a0
Correct Answer: D
Explanation:
Energy of photon $ = {{12500} \over {980}} = 12.75eV$
$ \therefore $ Electron will excite to n = 4
Since 'R' $ \propto $ n2
$ \therefore $ Radius of atom will be 16a0
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th January Evening Slot
Consider the nuclear fission
Ne20 $ \to $ 2He4 + C12
Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th January Morning Slot
Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to -
A.
200
B.
150
C.
400
D.
360
Correct Answer: A
Explanation:
at t = 0, A0 = ${{dN} \over {dt}}$ = 1600C/s
at t = 8s, A = 100 C/s
${A \over {{A_0}}}$ = ${1 \over {16}}$ in 8 sec
Therefor half life is t1/2 = 2 sec
$ \therefore $ Activity at t = 6 will be 1600${\left( {{1 \over 2}} \right)^3}$
= 200 C/s
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th January Evening Slot
At a given instant, say t = 0, two radioactive substance A and B have equal activities. the ratio ${{{R_B}} \over {{R_A}}}$ of their activities after time t itself decays with time t as e$-$3t. If the half-life of A is ln2, the half-life of B is :
$ \therefore $ Half life of B is $\left( {{t_{{1 \over 2}}}} \right) = {{\ln 2} \over {{\lambda _B}}} = {{\ln 2} \over 4}$
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th January Morning Slot
A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 $ \times $ 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive materail B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :
A.
5 days and 10 days
B.
10 days and 40 days
C.
20 days and 5 days
D.
20 days and 10 days
Correct Answer: C
Explanation:
Let number of nuclei present in material A is NA and in meterial B is NB According to question,
$ \therefore $ By checking options you can see possible value of ${\left( {{T_{{1 \over 2}}}} \right)_A}$ = 20 days and ${\left( {{T_{{1 \over 2}}}} \right)_B}$ = 5 days
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
At some instant, a radioactive sample S1 having an activity 5 $\mu $Ci has twice the number of nuclei as another sample S2 which has an activity of 10 $\mu $Ci. The half lives of S1 and S2 are :
A.
20 years and 5 years, respectively
B.
20 years and 10years, respectively
C.
5 years and 20 years, respectively
D.
10 years and 20 years, respectively
Correct Answer: A
Explanation:
As per question, N1
= 2N2
Also A1
= 5 $\mu $Ci, A2
= 10 $\mu $Ci
As A = $\lambda $N = ${{\ln 2} \over {{T_{1/2}}}}N$
So half life of sample S1 should be four timrs than sample S2.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let
${\lambda _n}$, ${\lambda _g}$ be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let
${\Lambda _n}$ be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)
A.
${\Lambda _n} \approx A + {B \over {\lambda _n^2}}$
B.
${\Lambda _n} \approx A + B{\lambda _n}$
C.
$\Lambda _n^2 \approx A + B\lambda _n^2$
D.
$\Lambda _n^2 \approx \lambda$
Correct Answer: A
Explanation:
We know,
Wavelength of emitted photon from n2 state to n1 state is
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its
energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Evening Slot
Muon ($\mu $$-$) is a negatively charged (|q| = |e|) particle with a mass m$\mu $ = 200 me, where me is the mass of the electron and e is the electronic charge. If $\mu $$-$ is bond to a proton to form a hydrogen like atom, identify the correct statements.
(A) Radis of the muonic orbit is 200 times smaller than that of the electron.
(B) The speed of the $\mu $$-$ in the nth orbit is ${1 \over {200}}$ times that of the electron in the nth orbit.
(C) The ionization energy of muonic atom is 200 timesmore than of an hydroen atom.
(D) The momentum of the muon in the nth orbit is 200 times more than that of the electron.
Only, m$\mu$ = 200 me rest are same. So, statement (C) is correct: The ionisation energy of muonic atom is 200 times more than that of an hydrogen atom.
(iv) Since Momentum $\propto$ Energy. So, statement (D) is correct: The momentum of the muon in he nth orbit is 200 times more than that of the electron.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Evening Slot
An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8 : 27. The ratio of the radii of the nuclei (assumed to be spherical) is :
A.
8 : 27
B.
4 : 9
C.
3 : 2
D.
2 : 3
Correct Answer: C
Explanation:
The two nuclei have velocity in ratio 8 : 27. By conservation of momentum, we have
Thus, ratio of radii of nuclei ${r_1}:{r_2} = 3:2$.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Morning Slot
The energy required to remove the electron from a singly ionized Helium atom is $2.2$ times the energies required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is :
A.
$20$ $eV$
B.
$34$ $eV$
C.
$79$ $eV$
D.
$109$ $eV$
Correct Answer: C
Explanation:
Energy required to remove e$-$ from singly ionized Helium atom
Let, E2 = energy required to remove e$-$from He $-$ atom
$\therefore\,\,\,\,$ According to q vertion.
E1 = 2.2 E2
$\therefore\,\,\,\,$ E2 = ${{54.4} \over {2.2}}$ = 24.72 eV
$\therefore\,\,\,\,$ Total energy required to ionize Helium atom completely
= (54.4 + 24.72) eV
= 79.12 eV
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Morning Slot
A solution containing active cobalt ${^{60}_{27}}Co$ having activity of $0.8$ $\mu Ci$ and decay constant $\lambda $ is injected in an animal's body. If $1\,c{m^3}$ of blood is drawn from the animal's body after $10$ hrs of injection, the activity found was $300$ decays per minute What is the volume of blood that is flowing in the body ? $\left( {\,\,Ci = 3.7 \times {{10}^{10}}\,} \right.$ decays per second and at $t=10$ hrs $\left. {{e^{ - \lambda t}} = 0.84} \right)$
A.
$6$ liters
B.
$7$ liters
C.
$4$ liters
D.
$5$ liters
Correct Answer: D
Explanation:
Initial activity, No = 0.8 $\mu $Ci
Activity at time t, N = N0 e$-$$\lambda $t
Activity in 1 cm3 blood after 10 hr, n = 300 decays per minute
= ${{300} \over {60}}\,dps$
= 5 dps.
Activity in whole blood after to hr = N0 e$-$$\lambda $$ \times $10
$\therefore\,\,\,\,$ Volume of the total blood = ${{{N_0}{e^{ - 10\lambda }}} \over n}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 9th April Morning Slot
Imagine that a reactor converts all given mass into energy and that it operates at a power level of 109 watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is
3×108 m/s)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 8th April Morning Slot
Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)
A.
30.2 MeV
B.
32.4 MeV
C.
23.6 MeV
D.
25.8 MeV
Correct Answer: C
Explanation:
1H2 + 1H2 $ \to $ 2He4
No. of proton in one dueteron = 2
$\therefore\,\,\,$ Total protons in two dueterons = 2 $ \times $ 2 = 4
$\therefore\,\,\,$ Binding energy of two dueteron
= 1.1 $ \times $ 4 = 4 : 4 MeV
In (2He4) no of protons = 4
$\therefore\,\,\,$ Binding energy of (2He4) nuclei = 4 $ \times $ 7 = 28 Mev
Energy released in this process = 28 $-$ 4.4 = 23.6 MeV
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 8th April Morning Slot
According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to : (n = principal quantum number)
A.
${n^{ - 4}}$
B.
${n^{ - 5}}$
C.
n$-$3
D.
n$-$2
Correct Answer: B
Explanation:
Magnetic field at the center of neucleus of H-atom
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Offline)
A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At
sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :
A.
$t = {T \over {\log (1.3)}}$
B.
$t = T\log (1.3)$
C.
$t = {T \over 2}{{\log 2} \over {\log 1.3}}$
D.
$t = T{{\log 1.3} \over {\log 2}}$
Correct Answer: D
Explanation:
Let initially there are total N0 number of nuclei.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Online) 9th April Morning Slot
A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :
A.
2
B.
3
C.
4
D.
5
Correct Answer: C
Explanation:
Energy released when hydrogen atom makes transition from n = 2 to n = 1 is,
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Offline)
Half-lives of two radioactive elements $A$ and $B$ are $20$ minutes and $40$ minutes, respectively. Initially, the samples have equal number of nuclei. After $80$ minutes, the ratio of decayed number of $A$ and $B$ nuclei will be:
A.
$1:4$
B.
$5:4$
C.
$1:16$
D.
$4:1$
Correct Answer: B
Explanation:
For ${A_{t{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} = 20\,\,$ min, $t=80$ min, number of half lifes $n=4$
$K.E = {k \over 2}{{z{e^2}} \over r}.$ Here $r$ decreases
2014
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2014 (Offline)
The radiation corresponding to $3 \to 2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field $3 \times {10^{ - 4}}\,T.$ If the radius of the larger circular path followed by these electrons is $10.0$ $mm$, the work function of the metal is close to:
A.
$1.8$ $eV$
B.
$1.1$ $eV$
C.
$0.8$ $eV$
D.
$1.6$ $eV$
Correct Answer: B
Explanation:
Radius of circular path followed by electron is given by,
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2014 (Offline)
Hydrogen $\left( {{}_1{H^1}} \right)$, Deuterium $\left( {{}_1{H^2}} \right)$, singly ionised Helium ${\left( {{}_2H{e^4}} \right)^ + }$ and doubly ionised lithium ${\left( {{}_3L{i^6}} \right)^{ + + }}$ all have one electron around the nucleus. Consider an electron transition from $n=2$ to $n=1.$ If the wavelengths of emitted radiation are ${\lambda _1},{\lambda _2},{\lambda _3}$ and ${\lambda _4}$ respectively then approximately which one of the following is correct?
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2013 (Offline)
In a hydrogen like atom electron make transition from an energy level with quantum number $n$ to another with quantum number $\left( {n - 1} \right)$. If $n > > 1,$ the frequency of radiation emitted is proportional to :
Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron $ = 1.6725 \times {10^{ - 27}}kg,$ mass of proton $ = 1.6725 \times {10^{ - 27}}\,kg,$ mass of electron $ = 9 \times {10^{ - 31}}\,kg$ ).
A diatomic molecule is made of two masses ${m_1}$ and ${m_2}$ which are separated by a distance $r.$ If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: ($n$ is an integer)
Hydrogen atom is excited from ground state to another state with principal quantum number equal to $4.$ Then the number of spectral lines in the emission spectra will be :
A.
$2$
B.
$3$
C.
$5$
D.
$6$
Correct Answer: D
Explanation:
The possible number of the spectral lines is given
The half life of a radioactive substance is $20$ minutes. The approximate time interval $\left( {{t_2} - {t_1}} \right)$ between the time ${{t_2}}$ when ${2 \over 3}$ of it had decayed and time ${{t_1}}$ when ${1 \over 3}$ of it had decayed is :
A.
$14$ min
B.
$20$ min
C.
$28$ min
D.
$7$ min
Correct Answer: B
Explanation:
Number of undecayed atom after time ${t_2};$
${{{N_0}} \over 3} = {N_0}{e^{ - \lambda {t_2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$
A nucleus of mass $M+$$\Delta m$ is at rest and decays into two daughter nuclei of equal mass ${M \over 2}$ each. Speed of light is $c.$
The binding energy per nucleon for the parent nucleus is ${E_1}$ and that for the daughter nuclei is ${E_2}.$ Then
A.
${E_2} = 2{E_1}$
B.
${E_1} > {E_2}$
C.
${E_2} > {E_1}$
D.
${E_1} = 2{E_2}$
Correct Answer: C
Explanation:
In nuclear fission, the binding energy per nucleon of daughter nuclei is greater than the parent nucleon of daughter nuclei is greater than the parent nucleus.
A radioactive nucleus (initial mass number $A$ and atomic number $Z$ emits $3\,\alpha $- particles and $2$ positrons. The ratio of number of neutrons to that of protons in the final nucleus will be
The transition from the state $n=4$ to $n=3$ in a hydrogen like atom result in ultra violet radiation. Infrared radiation will be obtained in the transition from :
A.
$3 \to 2$
B.
$4 \to 2$
C.
$5 \to 4$
D.
$2 \to 1$
Correct Answer: C
Explanation:
It is given that transition from the state $n=4$ to $n=3$ in a hydrogen like atom result in ultraviolet radiation. For infrared radiation the energy gap should be less. The only option is $5$ $ \to 4.$
The above is a plot of binding energy per nucleon ${E_b},$ against the nuclear mass $M;A,B,C,D,E,F$ correspond to different nuclei. Consider four reactions :
$\eqalign{
& \left( i \right)\,\,\,\,\,\,\,\,\,\,A + B \to C + \varepsilon \,\,\,\,\,\,\,\,\,\,\left( {ii} \right)\,\,\,\,\,\,\,\,\,\,C \to A + B + \varepsilon \,\,\,\,\,\,\,\,\,\, \cr
& \left( {iii} \right)\,\,\,\,\,\,D + E \to F + \varepsilon \,\,\,\,\,\,\,\,\,\,\left( {iv} \right)\,\,\,\,\,\,\,\,\,F \to D + E + \varepsilon ,\,\,\,\,\,\,\,\,\,\, \cr} $
where $\varepsilon $ is the energy released? In which reactions is $\varepsilon $ positive?
A.
$(i)$ and $(iii)$
B.
$(ii)$ and $(iv)$
C.
$(ii)$ and $(iii)$
D.
$(i)$ and $(iv)$
Correct Answer: D
Explanation:
For $A + B \to C + \varepsilon ,\,\,\varepsilon $ is positive. This is because ${E_b}$
for $C$ is greater than the ${E_b}$ for $A$ and $B$.
Again for $F \to D + E + \varepsilon ,\varepsilon $ is positive. This is
because ${E_b}$ for $D$ and $E$ is greater than ${E_b}$ for $F.$
This question contains Statement- 1 and Statement- 2. Of the four choices given after the statements, choose the one that best describes the two statements:
Statement- 1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and
Statement- 2: For heavy nuclei, binding energy per nucleon increases with increasing $Z$ while for light nuclei it decreases with increasing $Z.$
A.
Statement - $1$ is false, Statement - $2$ is true
B.
Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is a correct explanation for Statement - $1$
C.
Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is not a correct explanation for Statement - $1$
D.
Statement - $1$ is true, Statement - $2$ is false
Correct Answer: D
Explanation:
We know that energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Therefore statement $(1)$ is correct.
The second statement is false because for heavy nuclei the binding energy per nucleon decreases with increasing $Z$ and for light nuclei, B.E/nucleon increases with increasing $Z$ and for light nuclei, $B.E/$nucleon increases with increasing $Z.$
Suppose an electron is attracted towards the origin by a force ${k \over r}$ where $'k'$ is a constant and $'r'$ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the ${n^{th}}$ orbital of the electron is found to be $'{r_n}'$ and the kinetic energy of the electron to be $'{T_n}'.$