33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Nuclei
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123
29. (c) : 1 a.m.u = 931 MeV
30. (a) : Average binding energy/nucleon in nuclei is of
the order of 8 MeV.
31. (c) : Nuclear force is the same between any two
nucleons.
32. (c) : Nuclear forces are short range forces.
33. (b) : Alpha particle is a positively charged particle.
It is identical to the nucleus of the helium ( 2 He 4 ) atom,
so it contains 2 protons and 2 neutrons.
34. (d) : Given, t 1/2 = 2.2 × 10 9 s
and rate of radioactive disintegration,
dN
dt = −
10 10 s
1
0. 693 0.
693
∴ λ = = = 3.15 × 10 –10 s –1
t
9
22 . × 10
12 /
Now, we know that, N = N 0 e –lt
dN
−λt
⇒ =− λN0e
=−λN
dt
⇒ 10 10 = 3.15 × 10 –10 × N ⇒ N = 3.17 × 10 19
35. (a) : Number of nuclei remaining,
N = 600 – 450 = 150
According to the law of radioactive decay,
t
N 1 T12
= ⎛ N0
⎝ ⎜ ⎞ 2 ⎠ ⎟ /
; where N 0 is the number of nuclei initially.
t
150
∴ = ⎛ ⎝ ⎜ 1 ⎞
600 2 ⎠ ⎟
T 12 / ; where T 1/2 = half life.
t
2
⎛ 1 1
12
or
⎝
⎜ ⎞ 2 ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ 2 ⎠ ⎟ T /
⇒ t = 2T 1/2 = 2 × 10 minutes = 20 minutes
36. (*) : The number of radioactive nuclei ‘N’ at any
time t is given as
N(t) = N 0 e –lt
where N 0 is number of radioactive nuclei in the sample at
some arbitrary time t = 0 and l is the radioactive decay
constant.
Given: l A = 8l, l B = l, N 0A = N 0B = N 0
− λt
NB
e
∴ =
N
or 1 −λt 8λt 7λt
= e e = e
−8 λt
A e e
⇒ − 1=
7 = − 1
λt
or t
7λ
*Negative value of time is not possible.
N
So given ratio in question should be
B
= e
NA
37. (d) : N 0 = Nuclei at time t = 0
N 1 = Remaining nuclei after 40% decay
= (1 – 0.4) N 0 = 0.6 N 0
N 2 = Remaining nuclei after 85% decay
= (1 – 0.85) N 0 = 0.15 N 0
∴ = = = ⎛ ⎝ ⎜ ⎞ 2
N2
015 . N0
1 1
N N ⎠ ⎟
1 06 . 0 4 2
Hence, two half life is required between 40% decay and
85% decay of a radioactive substance.
\ Time taken = 2t 1/2 = 2 × 30 min = 60 min
38. (c) : If pTh
and pHe
are the momenta of thorium
and helium nuclei respectively, then according to law of
conservation of linear momentum
0 = p + p or p =−p
Th He Th He
Negative sign shows that both are moving in opposite
directions.
But in magnitude
p Th = p He
If m Th and m He are the masses of thorium and helium
nuclei respectively, then
2
p
Kinetic energy of thorium nucleus is Th
KTh
= and
that of helium nucleus is
2mTh
2
pHe
KTh
KHe
= ∴ = ⎛ p ⎞ ⎛
2 m
⎝ ⎜ Th mHe
⎞
K p ⎠
⎟
⎝
⎜ m ⎠
⎟
He He He Th
But p Th = p He and m He < m Th
\ K Th < K He or K He > K Th
Thus the helium nucleus has more kinetic energy than
the thorium nucleus.
39. (d) : Binding energy of 3
7 Li nucleus
= 7 × 5.60 MeV = 39.2 MeV
Binding energy of 2
4 He nucleus
= 4 × 7.06 MeV = 28.24 MeV
The reaction is
7 1
4
3Li + 1H
⎯→⎯ 2( 2He) + Q
∴ Q = 2( BE of 4 2He) − ( BE of 7
3Li)
= 2 × 28.24 MeV – 39.2 MeV
= 56.48 MeV – 39.2 MeV = 17.28 MeV
40. (c) : X → Y
Number of nuclei at t = 0 N 0 0
Number of nuclei after time t N 0 – x x
As per question, N 0 − x 1 =
x 7
7
7N0 − 7x = x or x=
N
8 0
\ Remaining nuclei of isotope X
= − = − = = ⎛ ⎝ ⎜ ⎞ 3
7 1 1
N0 x N0 N0 N0 ⎠ ⎟ N0
8 8 2
So three half lives would have been passed.
\ t = nT 1/2 = 3 × 1.4 × 10 9 years = 4.2 × 10 9 years
Hence, the age of the rock is 4.2 × 10 9 years.
41. (d) : There is requirement of three half lives so age
of the rock
t = nT 1/2 = 3 × 20 years = 60 years
42. (b) : For a given energy, g-rays has highest
penetrating power and a-particles has least penetrating
power.
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