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A radioactive isotope X with a half-life of $1.37 \times 10^{9}$ years decays to Y which is stable. A sample of rock from the moon was found to contain both the elements X and Y which were in the ratio of 1 : 7. The age of the rock is

$1.96 \times 10^{8}$ years

$3.85 \times 10^{9}$ years

$4.11 \times 10^{9}$ years

$9.59 \times 10^{9}$ years

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Explanation

(c) If in the rock there is no Y element, then the time taken by element X to reduce to $\frac{1}{8} th$ the initial value will be equal to $\frac{1}{8} = {(\frac{1}{2})}^{n}$ or n =3
Therefore, from the beginning three half life time is spent. Hence the age of the rock is

$= 3 \times 1.37 \times 10^{9} = 4.11 \times 10^{9} years$
.

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From a newly formed radioactive substance (Half life 2 hours), the intensity of radiation is 64 times the permissible safe level. The minimum time after which work can be done safely from this source is

6 hours

12 hours

24 hours

128 hours

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Explanation

(b) $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow \frac{1}{64} = {(\frac{1}{2})}^{6} = {(\frac{1}{2})}^{n} \Rightarrow n = 6$

After 6 half lives intensity emitted will be safe.
$∴$ Total time taken = $6 \times 2 = 12 hrs$

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The half life of radium is 1620 years and its atomic weight is 226 kgm per kilomol. The number of atoms that will decay from its 1 gm sample per second will be

$3.61 \times 10^{10}$

$3.6 \times 10^{12}$

$3.11 \times 10^{15}$

$31.1 \times 10^{15}$

(Avogardo's number N = $6.02 \times 10^{26}$ atpm/kilomol)

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Explanation

(a)

$\frac{dN}{dt} = λN; λ = \frac{0.6931}{t_{1 / 2}} = \frac{0.6931}{1620 \times 365 \times 24 \times 60 \times 60} N = \frac{6.023 \times 10^{23}}{226}$

$∴ \frac{dN}{dt} = \frac{0.6931 \times 6.023 \times 10^{23}}{1620 \times 365 \times 24 \times 60 \times 60 \times 226} = 3.61 \times 10^{10}$

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A radioactive material decays by simultaneous emission of two particles with respective half lives 1620 and 810 years. The time (in years) after which one- fourth of the material remains is

1080

2430

3240

4860

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Explanation

(a) $λ = λ_{1} + λ_{2} \Rightarrow \frac{\ln 2}{T} = \frac{\ln 2}{T_{1}} + \frac{\ln 2}{T_{2}}$

$∴ T = \frac{T_{1} T_{2}}{T_{1} + T_{2}} = \frac{810 \times 1620}{810 + 1620} = 540 years$

Hence $\frac{1}{4} th$ of material remain after 1080 years.

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The half life period of a radioactive element X is same as the mean life time of another radioactive element Y. Initially both of them have the same number of atoms. Then

X and Y have the same decay rate initially

X and Y decay at the same rate always

Y will decay at a faster rate than X

X will decay at a faster rate than Y

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Explanation

$\Rightarrow \frac{0.693}{λ_{x}} = \frac{1}{λ_{y}} \Rightarrow λ_{x} = 0.693 λ_{y} or λ_{x} < λ_{y}$

Also rate of decay = $λ$ N
Initially number of atoms (N) of both are equal but since $λ_{y} > λ_{x}$ therefore, y will decay at a faster rate than x.

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After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is

6000

9000

3000

24000

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Explanation

(d) Here the activity of the radioactive sample reduces to half in 140 days. Therefore, the half life of the sample is 140 days. 280 days is it’s two half lives. So before two half lives it’s activity was ( $2^{2} \times$ present activity).
$∴$ Initial activity = $2^{2} \times$ 6000=24000 dps

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Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is

54.4 eV

13.6 eV

40.8 eV

27.2 eV

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Explanation

(a) Excitation energy

$∆ E = E_{2} - E_{1} = 13.6 Z^{2} [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$

$\Rightarrow 40.8 = 13.6 \times \frac{3}{4} \times Z^{2} \Rightarrow Z = 2$

Now required energy to remove the electron from ground state

$= \frac{+ 13.6 Z^{2}}{1^{2}} = 13.6 Z^{2} = 54.4 eV$

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Consider a hydrogen like atom whose energy in nth exicited state is given by $E_{n} = - \frac{13.6 Z^{2}}{n^{2}}$ when this excited atom makes a transition from excited state to ground state, most energetic photons have energy $E_{\max}$ = 52.224 eV and least energetic photons have energy $E_{\min}$ = 1.224 eV. The atomic number of atom is

None of these

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Explanation

(a) Maximum energy is liberated for transition $E_{n} \to E_{1}$ and minimum energy for $E_{n} \to E_{n - 1}$
Hence $\frac{E_{1}}{n^{2}} - E_{1} = 52.224 eV$ ……(i)
and $\frac{E_{1}}{n^{2}} - \frac{E_{1}}{{(n - 1)}^{2}} = 1.224 eV$ …..(ii)
Solving equations (i) and (ii) we get
and $E_{1} = - 54.4 eV$ and n = 5
Now $E_{1} = - \frac{13.6 Z^{2}}{1^{2}} = - 54.4 eV$ . Hence Z = 2

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A radioactive sample is $α$ -emitter with half life 138.6 days is observed by a student to have 2000 disintegration/sec. The number of radioactive nuclei for given activity are

$3.45 \times 10^{10}$

$1 \times 10^{10}$

$3.45 \times 10^{15}$

$2.75 \times 10^{11}$

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Explanation

(a) Activity of substance that has 2000 disintegration/sec

The number of radioactive nuclei having activity A
$N = \frac{A}{λ} = \frac{2000 \times T_{1 / 2}}{\log_{e} 2} = \frac{2000 \times 138.6 \times 24 \times 3600}{0.693} = 3.45 \times 10^{10}$

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The ratio of radii of nuclei ${}_{13}^{27}{Al}$ and ${}_{52}^{A}X$ is 3 : 5. The number of neutrons in the nuclei of X will be

125

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Explanation

(b) $r \propto A^{1 / 3} \Rightarrow \frac{r_{1}}{r_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$

$\Rightarrow \frac{3}{5} = {(\frac{27}{A})}^{1 / 3} \Rightarrow \frac{27}{125} = \frac{27}{A} \Rightarrow A = 125$

Number of nuclei in atom X = A-52 = 125-52 = 73

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