33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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124 NEET-AIPMT Chapterwise Topicwise Solutions Physics
43. (d) : Let after t s amount of the A 1 and A 2 will
become equal in the mixture.
n
⎛ 1
As N = N
⎝
⎜
⎞ 0
2 ⎠ ⎟
where n is the number of half-lives
t/
20
⎛ 1
For A1,
N1 = N01
⎝
⎜
⎞ 2 ⎠ ⎟
t/
10
⎛ 1
For A2,
N2 = N02
⎝
⎜
⎞ 2 ⎠ ⎟
According to question, N 1 = N 2
⇒ 40 160
=
t/ 20 t/
10
2 2
2 t/10 = 4(2 t/20 ) or 2 t/10 = 2 2 2 t/20
⎛ t ⎞
+
⎝
⎜ 2
⎠
⎟ t t t t
t/ 10
⇒ 2 = 2 20 or = + 2 or − =2
10 20 10 20
t
or = 2 or t = 40 s
20
44. (b) : According to radioactive decay law
N = N 0 e –lt
where N 0 = Number of radioactive nuclei at time t = 0
N = Number of radioactive nuclei left undecayed at any
time t
l = decay constant
2
At time t 2 , of the sample had decayed
3
∴ N =
1 1
N
3 0 −λt
⇒ N = N e ...(i)
3 0 0 2
1
At time t 1 , of the sample had decayed,
3
∴ N =
2 N
3 0 ⇒ 2
= − t
N N e
λ ...(ii)
3 0 0 1
Divide (i) by (ii), we get
−λt
1 e 2 1 −λ
= ⇒ =
( t2−t
e
1)
⇒ l(t 2 – t 1 ) = ln2
2 −λt
e 1 2
ln2
t2 − t1 = ⇒ T12
/ =50 days
λ
n
45. (b) :
N 1
= ⎛ N ⎝ ⎜ ⎞ 0 2 ⎠ ⎟
where n is number of half lives
∴ = ⎛ ⎝ ⎜ ⎞ n
⎠ ⎟ ⎛
⎝
⎜ ⎞ 4
⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ n
1 1 1 1
or
⎠ ⎟ or n = 4
16 2 2 2
Let the age of rock be t years.
t
Q n =
T12
/
or t = nT 1/2 = 4 × 50 years = 200 years
46. (b) : Momentum of emitted photon
hυ
= pphoton
=
c
From the law of conservation of linear momentum,
hυ
p nucleus = p photon ∴ ⇒ Mv =
c
where v is the recoil speed of the nucleus
or
h
v = υ ...(i)
Mc
The recoil energy of the nucleus
2 2 2
1 2 1 ⎛ ⎞
= Mv = M
⎝
⎜
υ h υ
⎠
⎟ =
(Using(i))
2 2 Mc
2
2Mc
47. (c) : When an alpha particle 4
( 2 He ) is emitted, the
mass number and the atomic number of the daughter
nucleus decreases by four and two respectively. When a
beta particle (b – ) is emitted, the atomic number of the
daughter nucleus increases by one but the mass number
remains the same.
∴
− −
−
n m α m 4 2β
m 4
X ⎯ →⎯ n − 2Y ⎯⎯⎯→
n X
48. (c) : P Q
No. of nuclei, at t = 0 4N 0 N 0
Half- life 1 min 2 min
No. of nuclei after N P N Q
time t
Let after t min the number of nuclei of P and Q are equal.
⎛
∴ =
⎝
⎜ ⎞ t/ 1
⎠ ⎟ ⎛
N N N = N
⎝
⎜
⎞ t/2
1
1
P 4 0 and Q 0
2
2 ⎠ ⎟
As N P = N Q
∴
⎛
⎝
⎜ ⎞ t/ 1
1
⎠ ⎟ ⎛ 1
4 =
⎝
⎜
⎞ t/
2
t
2
N0
N0
2 2 ⎠ ⎟ or 4 =
t/
2 2
or 4 = 2 t/2 or 2 2 = 2 t/2 or
t
= 2 or t = 4min
2
After 4 minutes, both P and Q have equal number of
nuclei.
\ Number of nuclei of R
⎛ N0
⎞ ⎛ N ⎞
== −
⎝
⎜
⎠
⎟ + 0 15N 4N0
−
⎝
⎜N0
⎠
⎟ =
0 3N +
0 9N
=
0
4 4 4 4 2
49. (d) : According to activity law
R = R 0 e –lt ...(i)
According to given problem,
R 0 = N 0 counts per minute, R = N0 counts perminute
t = 5 minutes
e
Substituting these values in equation (i), we get
N0
−5λ = Ne or e
–1
= e –5l
0
e
1
5λ= 1or
λ=
perminute
5
At t = T 1/2 , the activity R reduces to R 0
2
where T 1/2 = half life of a radioactive sample
loge
2 loge
2
T12
/ = = 5loge
2
⎛ 1
⎝ ⎜ ⎞ = minutes
λ
5 ⎠ ⎟
50. (c) : A 1 = lN 1 at time t 1 , A 2 = lN 2 at time t 2
Therefore, number of nuclei decayed during time interval
(t 1 – t 2 ) is