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If the nuclear radius of ${}_{27}A l$ is 3.6 Fermi, the approximate nuclear radius of ${}_{64}C u$ in Fermi is

2.4

1.2

4.8

3.6

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Explanation

Nuclear radius $r \propto A^{1 / 3}$ , where A is mass number

r= $r_{0} A^{1 / 3}$

r= $r_{0} {(27)}^{1 / 3} = 3 r_{0}$

$r_{0} = \frac{3.6}{3} = 1.2 f m$

For ${}_{64}C u$

$r = r_{0} A^{1 / 3}$

=1.2fm ${(64)}^{1 / 3}$

=4.8 fm

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A mixture consists of two radioactive materials $A_{1}$ and $A_{2}$ with half lives of 20 s and 10 s respectively.Initially the mixture has 40g of $A_{1}$ and 160g of $A_{2}$ . The amount of the two in the mixture will become equal after

60 s

80 s

20 s

40 s

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Explanation

For 40 g amount

$40 g \to_{h a l f - l i f e}^{20 s} 20 g \overset{20 s}{\to} 10 g$

For 160g amount

$160 g \overset{10 s}{\to} 80 g \overset{10 s}{\to} 40 g$

$\overset{10 s}{\to} 20 g \overset{10 s}{\to} 10 g$

So, after 40s $A_{1} a n d A_{2}$ remains same.

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The half life of a radioactive nucleus is

50 days. The time interval $(t_{2} - t_{1})$ between

the time $t_{2}$ when $\frac{2}{3}$ of it has decayed and

the time $t_{1}$ when $\frac{1}{3}$ of it had decayed is

30 days

50 days

60 days

15 days

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Explanation

$A t t i m e t_{1}, \frac{N}{3} = N e^{- λ t 1} (1) A t t i m e t_{2}, \frac{2 N}{3} = N e^{- λ t 2} (2) (2) / (1) - 2 = \frac{e^{- λ t 2}}{e^{- λ t_{2}}} = e^{- λ t_{1} + λ t_{2}} - λ t_{1} + - λ t_{2} = \log 2 t_{2} - t_{1} = \frac{\log 2}{λ} = T_{\frac{1}{2}} = 50 d a y s$

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The power obtained in a reactor using $U^{235}$

disintegration is 1000 kW. The mass decay

of $U^{235}$ per hour is

$20 μg$

$40 μg$

$1 μg$

$10 μg$

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Explanation

Let assume power P=1000 W

Energy per hour= 1000 x 3600 J

Energy per fission=200 MeV

= $200 \times 1.6 \times 10^{- 13} J$

so, Number of fission per hour

n= $\frac{1000 \times 3600}{200 \times 1.6 \times 10^{- 13}}$

Number of mole per hour = $\frac{n}{N}$

so, Mass per hour= $\frac{n}{N} \times 235$

= $\frac{1000 \times 3600 \times 235}{200 \times 1.6 \times 10^{- 13} \times 6.02 \times 10^{23}} = 43.9 \times 10^{- 6} g$

This $43.9 \times 10^{- 6} g$ is nearest value of 40 micron

so option (b) is correct.

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The half-life of a radioactive isotope X is 50 yr.

It decays to another element Y which is stable.

The two elements X and Y were found to be in

the ratio of 1:15 in a sample of a given rock.

The age of rock was estimated to be

200 yr

250 yr

100 yr

150 yr

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Explanation

We know that

$\frac{N}{N_{0}} = {(\frac{1}{2})}^{t / t_{\frac{1}{2}}} \frac{1}{16} = {(\frac{1}{2})}^{t / 50} t = 4 \times 50 t = 200 yr$

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Fusion reaction takes place at high temperature because

atoms get ionised at high temperature

kinetic energy is high enough to overcome the Coulomb repulsion between nuclei

molecules break up at high temperature

nuclei break up at high temperature

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Explanation

Fusion reaction takes place at high temperatures because kinetic energy is high enough to overcome the Coulomb repulsion between nuclei.

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Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t=0, the number of P species are $4 N_{0}$ and that of Q is and that of Q are $N_{0} .$ Half-life of P(for conversion to R) is 1 min whereas that of Q is 2 min. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be:

$3 N_{0}$

$\frac{9 N_{0}}{2}$

$\frac{5 N_{0}}{2}$

$2 N_{0}$

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Explanation

Initially $P \to 4 N_{0}$

$Q \to N_{0}$

Half life $T_{P} \to 1 m i n$

$T_{Q} \to 2 m i n$

Let after time t number of nuclei of P and Q are equal i.e, $\frac{4 N_{0}}{2^{1 / 1}} = \frac{N_{0}}{2^{1 / 2}}$

$4 = 2^{1 / 2}$

$2^{2} = 2^{1 / 2}$

$\frac{t}{2} = 2$

t=4min

Disactive nucleus or Nuclei of R

$= (4 N_{0} - \frac{4 N_{0}}{2^{4}}) + (N_{0} - \frac{N_{0}}{2^{2}})$

$= 4 N_{0} - \frac{N_{0}}{4} + N_{0} - \frac{N_{0}}{4}$

$= 5 N_{0} - \frac{N_{0}}{2}$ $= \frac{9}{2} N_{0}$

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The mass of a ${}_{3}^{7}L i$ nucleus is 0.042u less than the sum of the masses of al its nucleons.The binding energy per nucleon of ${}_{3}^{7}L i$ nucleus is nearly

46 MeV

5.6 MeV

3.9 MeV

23 MeV

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Explanation

If m=1u, c= $3 \times 10^{8} m s^{- 1}, t h e n$

E=931 MeV ie, 1u=931 MeV

Binding energy=0.042 $\times$ 931=39.10 MeV

$∴$ Binding energy per nucleon

$= \frac{39.10}{7} = 5.58 = 5.6 M e V$

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The activity of a radioactive sample is measured as $N_{0}$ counts per minute at t=0 and $N_{0} / e$ counts per minute at t=5 min.The time (in minute) at which the activity reduces to half its value is

$\log_{e} 2 / 5$

$\frac{5}{\log_{e} 2}$

$5 \log_{10} 2$

$5 \log_{e} 2$

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Explanation

Fraction remains after n half lives

$\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} = {(\frac{1}{2})}^{t / T}$

Given $N = \frac{N_{0}}{e} \Rightarrow \frac{N_{0}}{e N_{0}} = {(\frac{1}{2})}^{5 / T}$

or $\frac{1}{e} = {(\frac{1}{2})}^{5 / T}$

Taking log on both sides, we get

$\log 1 - \log e = \frac{5}{T} \log \frac{1}{2}$

$- 1 = \frac{5}{T} (- \log 2)$

$\Rightarrow T = 5 \log_{e} 2$

Now, let $t^{'}$ be the time after which activity reduces to half

$(\frac{1}{2}) = {(\frac{1}{2})}^{t^{'} / 5 \log_{e} 2}$

$\Rightarrow t^{'} = 5 \log_{e} 2$

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The decay constant of a radio isotope is $λ .$ If $A_{1}$ and $A_{2}$ are its activities at times $t_{1}$ and $t_{2}$ respectively, the number of nuclei which have decayed during the time $(t_{1} - t_{2})$

$A_{1} t_{1} - A_{2} t_{2}$

$A_{1} - A_{2}$

$\frac{(A_{1} - A_{2})}{λ}$

$λ (A_{1} - A_{2})$

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Explanation

$A_{1} = λ N_{1} A_{2} = λ N_{2} N_{1} - N_{2} = [\frac{A_{1} - A_{2}}{λ}]$

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