Practice Start Test

Atoms and Nuclei

JEE Physics · 201 questions · Page 8 of 21 · Click an option or "Show Solution" to reveal answer

Q71
In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is λ\lambda . If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:
A 2516{{25} \over {16}} λ\lambda
B 2720{{27} \over {20}} λ\lambda
C 1625{{16} \over {25}} λ\lambda
D 2027{{20} \over {27}} λ\lambda
Correct Answer
Option D
Solution

For M →\to L steel

1λ{1 \over \lambda }

= K

(122−132)=K×536\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}

for N →\to L

1λ′{1 \over {\lambda '}}

= K

(122−142)=K×316\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}
λ′=2027λ\lambda ' = {{20} \over {27}}\lambda
Q72
A particle of mass m moves in a circular orbit in a central potential field U(r) = 12{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 n\sqrt n , En ∝ \propto n
B rn ∝ \propto n\sqrt n , En ∝ \propto 1n{1 \over n}
C rn ∝ \propto n, En ∝ \propto n
D rn ∝ \propto n2, En ∝ \propto 1n2{1 \over {{n^2}}}
Correct Answer
Option A
Solution

Force due to this field, F =

−∂U∂r- {{\partial U} \over {\partial r}}

F =

−∂∂r(12kr2)- {\partial \over {\partial r}}\left( {{1 \over 2}k{r^2}} \right)

= -kr For circular orbit,

mv2r{{m{v^2}} \over r}

= -kr ⇒\Rightarrow v ∝\propto r ..... (1) Fron Bohr’s quantization condition mvr =

nh2π{{nh} \over {2\pi }}

.....(2) From (1) and (2),

rn{r_n}

∝\propto

n12{n^{{1 \over 2}}}

Given, U(r) =

12{1 \over 2}

kr2 ⇒\Rightarrow En =

−12U(r)- {1 \over 2}U\left( r \right)

=

−14kr2- {1 \over 4}k{r^2}

⇒\Rightarrow En ∝\propto n

Q73
Given the masses of various atomic particles mp = 1.0072 u, mn = 1.0087 u, me = 0.000548 u, mv‾{m_{\overline v }} = 0, md = 2.0141 u, where p ≡ \equiv proton, n ≡ \equiv neutron, e ≡ \equiv electron, v‾\overline v ≡ \equiv antineutrino and d ≡ \equiv deuteron. Which of the following process is allowed by momentum and energy conservation?
A n + n → \to deuterium atom (electron bound to the nucleus)
B n + p → \to d + γ\gamma
C p → \to n + e+ + v‾\overline v
D e+ + e- → \to γ\gamma
Correct Answer
Option B
Solution

n + n →\to deuterium atom (This is incorrect) Correct is n + p →\to d + γ\gamma p →\to n + e+ +

v‾\overline v

(This is incorrect as mass is increasing) e+ + e- →\to γ\gamma (This is incorrect) Correct is e+ + e- →\to 2γ\gamma

Q74
You are given that Mass of 37Li{}_3^7Li = 7.0160u, Mass of 24He{}_2^4He = 4.0026u and Mass of 11H{}_1^1H = 1.0079u. When 20 g of 37Li{}_3^7Li is converted into 24He{}_2^4He by proton capture, the energy liberated, (in kWh), is : [Mass of nucleon = 1 GeV/c2]
A 6.82 × \times 105
B 4.5 × \times 105
C 8 × \times 106
D 1.33 × \times 106
Correct Answer
Option D
Solution
37Li{}_3^7Li

+

11H{}_1^1H

→\to 2(

24He{}_2^4He

)

Δ\Delta

m =

[MLi+MH]−2[MHe]\left[ {{M_{Li}} + {M_H}} \right] - 2\left[ {{M_{He}}} \right]

= (7.0160 + 1.0079) - 2 ×\times 4.0003 = 0.0187 Energy released in 1 reaction =

Δ\Delta

mc2 In use of 7.016 u Li energy is

Δ\Delta

mc2. In use of 1 gm Li energy is

Δmc2mLi{{\Delta m{c^2}} \over {{m_{Li}}}}

. In use of 20 gm energy is =

Δmc2mLi×20{{\Delta m{c^2}} \over {{m_{Li}}}} \times 20

=

0.087×1.6×10−19×1097.016×1.6×10−24×20{{0.087 \times 1.6 \times {{10}^{ - 19}} \times {{10}^9}} \over {7.016 \times 1.6 \times {{10}^{ - 24}}}} \times 20

J = 0.05 ×\times 1014 J = 1.33 ×\times 106 kWh

Q75
A radioactive nucleus decays by two different processes. The half life for the first process is 10 s and that for the second is 100 s. The effective half life of the nucleus is close to :
A 12 sec
B 9 sec
C 55 sec
D 6 sec
Correct Answer
Option B
Solution

T1 = 10 sec λ\lambda1 =

ln⁡2T1{{\ln 2} \over {{T_1}}}

T2 = 100 sec λ\lambda2 =

ln⁡2T2{{\ln 2} \over {{T_2}}}

λ\lambdaeq =

ln⁡2Teq{{\ln 2} \over {{T_{eq}}}}

We know, λ\lambdaeq = λ\lambda1 + λ\lambda2 ⇒\Rightarrow

ln⁡2Teq{{\ln 2} \over {{T_{eq}}}}

=

ln⁡2T1{{\ln 2} \over {{T_1}}}

+

ln⁡2T2{{\ln 2} \over {{T_2}}}

⇒\Rightarrow

1Teq{1 \over {{T_{eq}}}}

=

110+1100{1 \over {10}} + {1 \over {100}}

⇒\Rightarrow Teq =

10011{{100} \over {11}}

= 9 sec

Q76
Find the Binding energy per neucleon for 50120Sn{}_{50}^{120}Sn. Mass of proton mp = 1.00783 U, mass of neutron mn = 1.00867 U and mass of tin nucleus mSn = 119.902199 U. (take 1U = 931 MeV)
A 9.0 MeV
B 8.5 MeV
C 8.0 MeV
D 7.5 MeV
Correct Answer
Option B
Solution
B.E.=Δmc2B.E. = \Delta m{c^2}
=Δm×931= \Delta m \times 931
Δm=(50×1.00783)+(70×1.00867)−{119.902199}\Delta m = \left( {50 \times 1.00783} \right) + \left( {70 \times 1.00867} \right) - \left\{ {119.902199} \right\}
={120.9984−119.902199} U= \left\{ {120.9984 - 119.902199} \right\}\,U
=1.1238 U= 1.1238\,U
BE=1.1238 ×931=1046.2578 MeVBE = 1.1238\, \times 931 = 1046.2578\,MeV

BE per nucleon

≃\simeq

1046/120 ≈\approx 8.5 Mev

Q77
Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to :
A 2 : 1
B 1 : 2
C 5 : 7
D 10 : 7
Correct Answer
Option A
Solution

We know, kinetic energy K = qV also

K=P22mK = {{{P^2}} \over {2m}}

∴\therefore

qV=P22m=m2v22mqV = {{{P^2}} \over {2m}} = {{{m^2}{v^2}} \over {2m}}
V=2qvmV = \sqrt {{{2qv} \over m}}
V∝qmV \propto \sqrt {{q \over m}}
VHVHe=eme4m=21{{{V_H}} \over {{V_{He}}}} = {{\sqrt {{e \over m}} } \over {\sqrt {{e \over {4m}}} }} = {2 \over 1}
Q78
In a radioactive material, fraction of active material remaining after time t is 9/16. The fraction that was remaining after t/2 is
A 34{3 \over 4}
B 45{4 \over 5}
C 35{3 \over 5}
D 78{7 \over 8}
Correct Answer
Option A
Solution

First order decay N(t) = N0e-λ\lambdat Given

N(t)N0=916={{N\left( t \right)} \over {{N_0}}} = {9 \over {16}} =

e-λ\lambdat N(t/2) = N0e-λ\lambda(t/2)

N(t/2)N0=e−λt{{N\left( {t/2} \right)} \over {{N_0}}} = \sqrt {{e^{ - \lambda t}}}

=

916\sqrt {{9 \over {16}}}

=

34{3 \over 4}

⇒\Rightarrow

N(t/2)=34N0N\left( {t/2} \right) = {3 \over 4}{N_0}
Q79
The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × \times 10-16 s. The frequency of revolution of the electron in its first excited state (in s-1) is :
A 5.6 × \times 1012
B 1.6 × \times 1014
C 7.8 × \times 1014
D 6.2 × \times 1015
Correct Answer
Option C
Solution

Time period of revolution of electron in nth orbit V =

2πrV{{2\pi r} \over V}

=

2πa0(n2Z)V0(Zn){{2\pi {a_0}\left( {{{{n^2}} \over Z}} \right)} \over {{V_0}\left( {{Z \over n}} \right)}}

⇒\Rightarrow T ∝\propto

n3Z2{{{{n^3}} \over {{Z^2}}}}

∴\therefore

T1T2=n13n23{{{T_1}} \over {{T_2}}} = {{n_1^3} \over {n_2^3}}

⇒\Rightarrow

1.6×10−16T2=1(2)3{{1.6 \times {{10}^{ - 16}}} \over {{T_2}}} = {1 \over {{{\left( 2 \right)}^3}}}

⇒\Rightarrow T2 = 1.6 ×\times 8 ×\times 10-16 ∴\therefore f2 =

1T2{1 \over {{T_2}}}

=

11.6×8×10−16{1 \over {1.6 \times 8 \times {{10}^{ - 16}}}}

= 7.8 ×\times 1014

Q80
The activity of a radioactive sample falls from 700 s–1 to 500 s–1 in 30 minutes. Its half life is close to:
A 62 min
B 66 min
C 72 min
D 52 min
Correct Answer
Option A
Solution

A = A0 e-λ\lambdat ⇒\Rightarrow 500 = 700 e-λ\lambda×\times30 ⇒\Rightarrow

ln⁡75\ln {7 \over 5}

= λ\lambda×\times30 Also T1/2 =

ln⁡2λ{{\ln 2} \over \lambda }

=

ln⁡2ln⁡75×30{{\ln 2} \over {\ln {7 \over 5}}} \times 30

⇒\Rightarrow T1/2 = 62 min

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