46: The Nucleus : Neet-xii-physics

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NEET-XII-Physics

46: The Nucleus

with Solutions - page 7

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Qstn #27
The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one month old ³²P(t₁_/2 = 14.3 days) source if it was originally purchased for 800 rupees?
Ans : Given:
Half-life of ³²P source, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}``= 14.3 days
Time, t = 30 days = 1 month
Here, the selling rate of a radioactive isotope is decided by its activity.
∴ Selling rate = Activity of the radioactive isotope after 1 month
Initial activity, A₀ = 800 disintegration/sec
Disintegration constant `` \left(\lambda \right)`` is given by
`` \lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}`` = `` \frac{0.693}{14.3}{\,\mathrm{\,days\,}}^{-1}``
Activity `` \left(A\right)`` is given by
A = A₀e^-^λt
Here, `` \lambda `` = Disintegration constant
`` \therefore `` Activity of the radioactive isotope after one month (selling rate of the radioactive isotope) `` \left(A\right)`` is given below.
`` A=800\times {e}^{\frac{-0.693}{14.3}\times 30}``
`` =800\times 0.233669``
`` =186.935=\,\mathrm{\,Rs\,}187``
Page No 443:

Qstn #28
⁵⁷Co decays to ⁵⁷Fe by β⁺- emission. The resulting ⁵⁷Fe is in its excited state and comes to the ground state by emitting γ-rays. The half-life of β⁺- decay is 270 days and that of the γ-emissions is 10^-8 s. A sample of ⁵⁷Co gives 5.0 × 10⁹ gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5 × 10⁹ per second?
Ans : According to the question, when the β⁺ decays to half of its original amount, the emission rate of γ-rays will drop to half. For this, the sample will take 270 days.
Therefore, the required time is 270 days.
Page No 443:

Qstn #29
Carbon (Z = 6) with mass number 11 decays to boron (Z = 5).

#29-a
Is it a β⁺-decay or a β^-decay?
Ans : The reaction is given by
`` {\,\mathrm{\,C\,}}_{6}\to {\,\mathrm{\,B\,}}_{5}+{\,\mathrm{\,\beta \,}}^{+}+\,\mathrm{\,v\,}``
It is a β⁺ decay since atomic number is reduced by 1.

#29-b
The half-life of the decay scheme is 20.3 minutes. How much time will elapse before a mixture of 90% carbon-11 and 10% boron-11 (by the number of atoms) converts itself into a mixture of 10% carbon-11 and 90% boron-11?
Ans : Half-life of the decay scheme, T _1/2 = 20.3 minutes
Disintegration constant, `` \lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}`` = `` \frac{0.693}{20.3}{\,\mathrm{\,min\,}}^{-1}``
If t is the time taken by the mixture in converting, let the total no. of atoms be 100N₀_.

Carbon Boron

Initial 90 N₀ 10 N₀

Final 10 N₀ 90 N₀

N = N₀e^-λ^t
Here, N₀ = Initial number of atoms
N = Number of atoms left undecayed
10N₀ =90N₀e^-λ^t ( For carbon)
`` \Rightarrow \frac{1}{9}={e}^{-\frac{0.693}{20.3}\times t}``
`` \Rightarrow \,\mathrm{\,In\,}\frac{1}{9}=\frac{-0.693}{20.3}t``
`` \Rightarrow t=64.36=64\,\mathrm{\,min\,}``
Page No 443:

Qstn #30
4 × 10²³ tritium atoms are contained in a vessel. The half-life of decay tritium nuclei is 12.3 y. Find (a) the activity of the sample, (b) the number of decay in the next 10 hours (c) the number of decays in the next 6.15 y.
digAnsr: b
Ans : Given:
Number of tritium atoms, N₀ = 4 × 10²³
Half-life of tritium nuclei, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}``= 12.3 years
Disintegration constant, `` \lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{0.693}{12.3}`` years^{`` -1``}
Activity of the sample `` \left(A\right)`` is given by
A₀ = `` \frac{dN}{dt}`` = `` \lambda {N}_{0}``
`` \Rightarrow {A}_{0}=\frac{0.693}{{t}_{1/2}}{N}_{0}``
`` =\frac{0.693}{12.3}\times 4\times {10}^{23}\,\mathrm{\,disintegration\,}/\,\mathrm{\,year\,}``
`` =\frac{0.693\times 4\times {10}^{23}}{12.3\times 3600\times 24\times 365}\,\mathrm{\,disintegration\,}/\,\mathrm{\,sec\,}``
`` =7.146\times {10}^{14}\,\mathrm{\,disintegration\,}/\,\mathrm{\,sec\,}``
`` ``
(b) Activity of the sample, A = 7.146`` \times ``10¹⁴ disintegration/sec
`` \,\mathrm{\,Number\,}\,\mathrm{\,of\,}\,\mathrm{\,decays\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,next\,}10\,\mathrm{\,h\,}\,\mathrm{\,ours\,}=7.146\times {10}^{14}\times 10\times 3600``
`` =257.256\times {10}^{17}``
`` =2.57\times {10}^{19}``
`` ``
(c) Number of atoms left undecayed, `` N={N}_{0}{e}^{-\lambda t}``
Here, N₀ = Initial number of atoms
`` \therefore N=4\times {10}^{23}\times {e}^{\frac{-0.693}{12.3}\times 6.15}=2.83\times {10}^{23}``
Number of atoms disintegrated = (N₀_{`` -``}N) = (4`` -``2.83) × 10²³ = 1.17 × 10²³
Page No 443:

Qstn #31
A point source emitting alpha particles is placed at a distance of 1 m from a counter which records any alpha particle falling on its 1 cm² window. If the source contains 6.0 × 10¹⁶ active nuclei and the counter records a rate of 50000 counts/second, find the decay constant. Assume that the source emits alpha particles uniformly in all directions and the alpha particles fall nearly normally on the window.
Ans : Given:
Counts received per second = 50000 Counts/second
Number of active nuclei, N = 6 × 10¹⁶
Total counts radiated from the source, `` \frac{dN}{\,\mathrm{\,dt\,}}`` = Total surface area × 50000 counts/cm²
= 4 × 3.14 × 1 × 10⁴ × 5 × 10⁴
= 6.28 × 10⁹ Counts
We know
`` \frac{dN}{dt}=\lambda N``
Here, λ = Disintegration constant
`` \therefore \lambda =\frac{6.28\times {10}^{9}}{6\times {10}^{16}}``
`` =1.0467\times {10}^{-7}``
`` =1.05\times {10}^{-7}{\,\mathrm{\,s\,}}^{-1}``
Page No 443:

Qstn #32
²³⁸U decays to ²⁰⁶Pb with a half-life of 4.47 × 10⁹ y. This happens in a number of steps. Can you justify a single half for this chain of processes? A sample of rock is found to contain 2.00 mg of ²³⁸U and 0.600 mg of ²⁰⁶Pb. Assuming that all the lead has come from uranium, find the life of the rock.
Ans : Given:
Half-life of ²³⁸U, t₁_/2 = 4.47 × 10⁹ years
`` \,\mathrm{\,Total\,}\,\mathrm{\,number\,}\,\mathrm{\,of\,}\,\mathrm{\,atoms\,}\,\mathrm{\,present\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,rock\,}initially,{N}_{0}=\left(\frac{6.023\times {10}^{23}\times 2}{238}+\frac{6.023\times {10}^{23}\times 0.6}{206}\right)``
`` =\left(\frac{12.046}{238}+\frac{3.62}{206}\right)\times {10}^{20}``
`` =\left(0.0505+0.0175\right)\times {10}^{20}``
`` =0.0680\times {10}^{20}``
`` ``
Now, N = N₀e^{`` -\lambda t``}
Here, `` \lambda `` = Disintegration constant
t = Life of the rock
`` \Rightarrow N={N}_{0}{e}^{\frac{-0.693}{{t}_{1/2}}\times t}``
`` \Rightarrow 0.0505=0.0680{e}^{\frac{-0.693}{4.47\times {10}^{9}}\times t}``
`` \Rightarrow \,\mathrm{\,ln\,}\left(\frac{0.0505}{0.0680}\right)=\frac{-0.6931}{4.47\times {10}^{9}}\times t``
`` \Rightarrow t=1.92\times {10}^{9}\,\mathrm{\,years\,}``
Page No 443:

Qstn #33
When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of ¹⁴C per gram per minute. A sample from an ancient piece of charcoal shows ¹⁴C activity to be 12.3 disintegrations per gram per minute. How old is this sample? Half-life of ¹⁴C is 5730 y.
Ans : Given:
Initial activity of charcoal, A₀ = 15.3 disintegrations per gram per minute
Half-life of charcoal, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}``= 5730 years
Final activity of charcoal after a few years, A = 12.3 disintegrations per gram per minute
Disintegration constant, `` \lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}`` = `` \frac{0.693}{5370}{\,\mathrm{\,y\,}}^{-1}``
Let the sample take a time of t years for the activity to reach 12.3 disintegrations per gram per minute.
Activity of the sample, `` A={A}_{0}{e}^{-\lambda t}``
`` A={A}_{0}{e}^{-\frac{0.693}{5730}\times t}``
`` \Rightarrow \,\mathrm{\,In\,}\frac{12.3}{15.3}=\frac{-0.693}{5730}t``
`` \Rightarrow 0.218253=\frac{0.693}{5730}\times t``
`` \Rightarrow t=1804.3\,\mathrm{\,years\,}``
Page No 443:

Qstn #34
Natural water contains a small amount of tritium (
H13). This isotope beta-decays with a half-life of 12.5 years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things he finds a sealed bottled of whisky. On returning, he analyses the whisky and finds that it contains only 1.5 per cent of the
H13radioactivity as compared to a recently purchased bottle marked ‘8 years old’. Estimate the time of that unsuccessful attempt.
Ans : Given:
Half-life time of tritium, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}`` = 12.5 years
Disintegration constant, `` \lambda =\frac{0.693}{12.5}\,\mathrm{\,per\,}\,\mathrm{\,year\,}``
Let A₀ be the activity, when the bottle was manufactured.
Activity after 8 years `` \left(A\right)`` is given by
`` A={A}_{0}{e}^{\frac{-0.693}{12.5}\times 8}`` ...(1)
Let us consider that the mountaineering had taken place t years ago.
Then, activity of the bottle `` \left(A\text{'}\right)`` on the mountain is given by
`` A\text{'}={A}_{0}{e}^{-\lambda t}``
Here, A' = (Activity of the bottle manufactured 8 years ago) × 1.5%
`` A\text{'}={A}_{0}{e}^{\frac{-0.693}{12.5}\times 8}\times 0.015...\left(2\right)``
`` ``
Comparing (1) and (2)
`` \frac{0.693}{12.5}t=\frac{-0.6931\times 8}{12.5}+\,\mathrm{\,In\,}[0.015]``
`` \Rightarrow \frac{-0.693}{12.5}t=\frac{-0.693}{12.5}\times 8-4.1997``
`` \Rightarrow 0.693t=58.040``
`` \Rightarrow t=83.75\,\mathrm{\,years\,}``
Page No 443:

Qstn #35
The count rate of nuclear radiation coming from a radiation coming from a radioactive sample containing ¹²⁸I varies with time as follows.

Time t (minute): 0 25 50 75 100

Ctount rate R (10⁹ s^-1): 30 16 8.0 3.8 2.0

#35-a
Plot In (R₀/R) against t.
Ans : For t = 0,
`` \,\mathrm{\,ln\,}\left(\frac{{R}_{\mathit{0}}}{R}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{30\times {10}^{9}}\right)=0``
`` ``
For t = 25 s,
`` \,\mathrm{\,ln\,}\left(\frac{{R}_{\mathit{0}}}{{R}_{\mathit{2}}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{16\times {10}^{9}}\right)=0.63``
For t = 50 s,
`` \,\mathrm{\,In\,}\left(\frac{{\mathit{R}}_{\mathit{0}}}{{\mathit{R}}_{\mathit{3}}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{8\times {10}^{9}}\right)=1.35``
For t = 75 s,
`` \,\mathrm{\,ln\,}\left(\frac{{R}_{0}}{{R}_{4}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{3.8\times {10}^{9}}\right)=2.06``
For t = 100 s,
`` \,\mathrm{\,In\,}\left(\frac{{R}_{0}}{{R}_{5}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{2\times {10}^{9}}\right)=2.7``
The required graph is shown below.

#35-b
From the slope of the best straight line through the points, find the decay constant λ. (c) Calculate the half-life t₁_/2.
Ans : Slope of the graph = 0.028
∴ Decay constant, `` \lambda `` = 0.028 min^{`` -1``}
The half-life period`` \left({T}_{\frac{1}{2}}\right)`` is given by
`` {T}_{\frac{1}{2}}=\frac{0.693}{\lambda }``
`` =\frac{0.693}{0.028}=25\,\mathrm{\,min\,}``
Page No 443:

Qstn #36
The half-life of ⁴⁰K is 1.30 × 10⁹ y. A sample of 1.00 g of pure KCI gives 160 counts s^-1. Calculate the relative abundance of ⁴⁰K (fraction of ⁴⁰K present) in natural potassium.
Ans : Given:
Half-life period of ⁴⁰K, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}`` = 1.30 × 10⁹ years
Count given by 1 g of pure KCI, A = 160 counts/s
Disintegration constant, `` \lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}``
Now, activity, A = λN
`` \Rightarrow 160=\frac{0.693}{{t}_{1/2}}\times N``
`` \Rightarrow 160=\left(\frac{0.693}{1.30\times {10}^{9}\times 365\times 86400}\right)\times N``
`` \Rightarrow N=\frac{160\times 1.30\times 365\times 86400\times {10}^{9}}{0.693}``
`` \Rightarrow N=9.5\times {10}^{18}``
6.023 × 10²³ atoms are present in 40 gm.
`` \,\mathrm{\,Thus,\,}\,\mathrm{\,9.5\; \times \; 1018\,}\,\mathrm{\,atoms\,}\,\mathrm{\,will\,}\,\mathrm{\,be\,}\,\mathrm{\,present\,}in\frac{40\times 9.5\times {10}^{18}}{6.023\times {10}^{23}}\,\mathrm{\,gm\,}.``
`` =\frac{4\times 9.5\times {10}^{-4}}{6.023}\,\mathrm{\,gm\,}``
`` =6.309\times {10}^{-4}``
`` =0.00063\,\mathrm{\,gm\,}``
Relative abundance of ⁴⁰K in natural potassium = (2 × 0.00063 × 100)% = 0.12%
Page No 443:

Qstn #37
Hg80197decay to
Au79197through electron capture with a decay constant of 0.257 per day.
Ans : Given:
Decay constant of electron capture = 0.257 per day

Time t (minute):	0	25	50	75	100
Ctount rate R (10⁹ s^-1):	30	16	8.0	3.8	2.0

	Carbon	Boron
Initial	90 N₀	10 N₀
Final	10 N₀	90 N₀