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Atoms and Nuclei

JEE Physics · 201 questions · Page 12 of 21 · Click an option or "Show Solution" to reveal answer

Q111
A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of : (Given, Planck's constant = 6.6 ×\times 10−-34 Js).
A 2.10 ×\times 10−-34 Js
B 1.05 ×\times 10−-34 Js
C 3.15 ×\times 10−-34 Js
D 4.2 ×\times 10−-34 Js
Correct Answer
Option B
Solution
−13.6+10.2=−13.6n2- 13.6 + 10.2 = {{ - 13.6} \over {{n^2}}}
⇒13.6n2=3.4\Rightarrow {{13.6} \over {{n^2}}} = 3.4
⇒n=2\Rightarrow n = 2
⇒ΔL=2×h2λ−1×h2λ\Rightarrow \Delta L = 2 \times {h \over {2\lambda }} - 1 \times {h \over {2\lambda }}
=h2λ= {h \over {2\lambda }}
⇒ΔL≃1.05×10−34\Rightarrow \Delta L \simeq 1.05 \times {10^{ - 34}}

Js

Q112
How many alpha and beta particles are emitted when Uranium 92U238 decays to lead 82Pb206 ?
A 3 alpha particles and 5 beta particles
B 6 alpha particles and 4 beta particles
C 4 alpha particles and 5 beta particles
D 8 alpha particles and 6 beta particles
Correct Answer
Option D
Solution
92U238⟶82Pb206+x(2He4)+y(−1β0){}_{92}{U^{238}}\overset{{}}\longrightarrow {}_{82}P{b^{206}} + x\left( {{}_2H{e^4}} \right) + {}_y\left( {{}_{ - 1}{\beta ^0}} \right)
238=206+4x+0238 = 206 + 4x + 0
⇒4x=32⇒x=8\Rightarrow 4x = 32 \Rightarrow x = 8

also,

92=82+2x−y92 = 82 + 2x - y
y=82+16−92=6y = 82 + 16 - 92 = 6
Q113
A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:
A 3.75 hours
B 0.56 hours
C 0.26 hours
D 1.80 hours
Correct Answer
Option D
Solution
dAdt−(−λ1A)+(−λ2A){{dA} \over {dt}} - ( - {\lambda _1}A) + ( - {\lambda _2}A)
⇒dAdt=−(λ1+λ2)A\Rightarrow {{dA} \over {dt}} = - ({\lambda _1} + {\lambda _2})A
⇒λeff=λ1+λ2\Rightarrow {\lambda _{eff}} = {\lambda _1} + {\lambda _2}
⇒ln⁡2(t1/2)eff=ln⁡2(t1/2)1+ln⁡2(t1/2)2\Rightarrow {{\ln 2} \over {{{({t_{1/2}})}_{eff}}}} = {{\ln 2} \over {{{({t_{1/2}})}_1}}} + {{\ln 2} \over {{{({t_{1/2}})}_2}}}
⇒(t1/2)eff=4.5×37.5\Rightarrow {({t_{1/2}})_{eff}} = {{4.5 \times 3} \over {7.5}}

hours = 1.8 hours

Q114
In Bohr's atomic model of hydrogen, let K, P and E are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level :
A All K, P and E increase.
B K decreases, P and E increase.
C P decreases, K and E increase.
D K increases, P and E decrease.
Correct Answer
Option B
Solution
T.E.=−Z2me48(nhε0)2T.E. = {{ - {Z^2}m{e^4}} \over {8{{\left( {nh{\varepsilon _0}} \right)}^2}}}
P.E.=−Z2me44(nhε0)2P.E. = {{ - {Z^2}m{e^4}} \over {4{{\left( {nh{\varepsilon _0}} \right)}^2}}}
K.E.=Z2me48(nhε0)2K.E. = {{{Z^2}m{e^4}} \over {8{{\left( {nh{\varepsilon _0}} \right)}^2}}}

As electron makes transition to higher level, total energy and potential energy increases (due to negative sign) while the kinetic energy reduces.

Q115
Choose the correct option from the following options given below :
A In the ground state of Rutherford's model electrons are in stable equilibrium. While in Thomson's model electrons always experience a net-force.
B An atom has a nearly continuous mass distribution in a Rutherford's model but has a highly non-uniform mass distribution in Thomson's model.
C A classical atom based on Rutherford's model is doomed to collapse.
D The positively charged part of the atom possesses most of the mass in Rutherford's model but not in Thomson's model.
Correct Answer
Option C
Solution

An atom based on classical theory of Rutherford's model should collapse as the electrons in continuous circular motion that is a continuously accelerated charge should emit EM waves and so should lose energy.

These electrons losing energy should soon fall into heavy nucleus collapsing the whole atom.

Q116
Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments 'B' and 'C' of mass numbers 105 and 115. The binding energy of nucleons in 'B' and 'C' is 6.4 MeV per nucleon. The energy Q released per fission will be :
A 0.8 MeV
B 275 MeV
C 220 MeV
D 176 MeV
Correct Answer
Option D
Solution

220A →\to 105B + 115C ⇒\Rightarrow Q = [105 ×\times 6.4 + 115 ×\times 6.4] −- [220 ×\times 5.6] MeV ⇒\Rightarrow Q = 176 MeV

Q117
A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :
A 1.05 ×\times 10−-34 Js
B 2.11 ×\times 10−-34 Js
C 3.16 ×\times 10−-34 Js
D 4.22 ×\times 10−-34 Js
Correct Answer
Option B
Solution

Change in energy

ΔE=Ef−E2⇒12.09=Ef−(−13.6)⇒Ef=−1.51eV⇒−13.6n2=−1.51⇒n2=9⇒n=3 So, ΔL=Lf−Li=h2π(3−1)=2h2π=hπ=6.63×10−343.14=2.11×10−34Js\begin{aligned} &\Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{2} \Rightarrow 12.09=\mathrm{E}_{\mathrm{f}}-(-13.6) \Rightarrow \mathrm{E}_{\mathrm{f}}=-1.51 \mathrm{eV} \\\\ &\Rightarrow \frac{-13.6}{n^{2}}=-1.51 \Rightarrow \mathrm{n}^{2}=9 \Rightarrow \mathrm{n}=3 \\\\ &\text{ So, } \Delta \mathrm{L}=\mathrm{L}_{\mathrm{f}}-\mathrm{L}_{\mathrm{i}} \\\\ &=\frac{h}{2 \pi}(3-1)=\frac{2 h}{2 \pi}=\frac{h}{\pi}=\frac{6.63 \times 10^{-34}}{3.14}=2.11 \times 10^{-34} \mathrm{Js} \end{aligned}
Q118
The momentum of an electron revolving in nth \mathrm{n}^{\text{th }} orbit is given by : (Symbols have their usual meanings)
A nh2πr\dfrac{\mathrm{nh}}{2 \pi \mathrm{r}}
B nh2r \dfrac{n h}{2 r}
C nh2π \dfrac{\mathrm{nh}}{2 \pi}
D 2πrnh\dfrac{2 \pi r}{\mathrm{nh}}
Correct Answer
Option A
Solution

∵\because

mvr=nh2πmvr = {{nh} \over {2\pi }}
⇒mv=nh2πr\Rightarrow mv = {{nh} \over {2\pi r}}
Q119
Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength λ\lambda. The value of principal quantum number 'nn' of the excited state will be : (R:\mathrm{R}: Rydberg constant)
A λRλ−1\sqrt{\dfrac{\lambda \mathrm{R}}{\lambda-1}}
B λRλR−1\sqrt{\dfrac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}
C λλR−1\sqrt{\dfrac{\lambda}{\lambda \mathrm{R}-1}}
D λR2λR−1\sqrt{\dfrac{\lambda R^{2}}{\lambda R-1}}
Correct Answer
Option B
Solution

∵\because

1λ=R(112−1n2){1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)
⇒1λR=1−1n2\Rightarrow {1 \over {\lambda R}} = 1 - {1 \over {{n^2}}}
⇒1n2=1−1λR=λR−1λR\Rightarrow {1 \over {{n^2}}} = 1 - {1 \over {\lambda R}} = {{\lambda R - 1} \over {\lambda R}}
⇒n=λRλR−1\Rightarrow n = \sqrt {{{\lambda R} \over {\lambda R - 1}}}
Q120
The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is : (\left(\right.Take log⁡101.88=0.274)\left.\log _{10} 1.88=0.274\right)
A 0.02min⁡−10.02 \min ^{-1}
B 2.7min⁡−12.7 \min ^{-1}
C 0.063min⁡−10.063 \min ^{-1}
D 6.3min⁡−16.3 \min ^{-1}
Correct Answer
Option C
Solution
A0=4250{A_0} = 4250
A=2250=A0e−λtA = 2250 = {A_0}{e^{ - \lambda t}}
⇒22504250=e−λt\Rightarrow {{2250} \over {4250}} = {e^{ - \lambda t}}
⇒λ(10)=ln⁡(42502250)\Rightarrow \lambda (10) = \ln \left( {{{4250} \over {2250}}} \right)
λ(10)=0.636\lambda (10) = 0.636
λ=0.063\lambda = 0.063
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