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An archeologist analyses the wood in a prehistoric structure and finds that $C^{14}$ (Half - life=5700 years) to $C^{12}$ is only one - fourth of that found in the cells buried plants. The age of the wood is about

5700 yr

2850 yr

11,400 yr

22,800 yr

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Explanation

In radioactive dating, the age of an object is determined by measuring the remaining amount of a radioactive isotope. If the amount of $C^{14}$ is one-fourth of the initial amount, it means that three half-lives have passed. So, the age of the wood is 3 Ã— 5700 = 11,400 years.

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In the nuclear reaction: $X {(n, α)}_{3} L i^{7}$ the term X will be :

${}_{5}B_{10}$

${}_{5}B_{9}$

${}_{5}B_{11}$

${}_{2}H e^{4}$

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Explanation

$X (n, α) {}_{3}^{7}L i \Rightarrow {}_{Z}X_{A} + {}_{0}n_{1} \to {}_{3}L i^{7} + {}_{2}H e^{4} Z = 3 + 2 = 5 a n d A = 7 + 4 - 1 = 10 ∴ {}_{5}X_{10} = {}_{5}B_{10}$

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10 g of radioactive material of half-life 15 year is kept in store for 20 years. The disintegrated material mass is [Pb. PMT 2002]

12.5 g

10.5 g

6.03 g

4.03 g

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Explanation

Remaining material $N = \frac{N_{0}}{2^{t / T}}$

$\Rightarrow N = \frac{10}{{(2)}^{20 / 15}} = \frac{10}{2.51} = 3.96 g$

So decayed material = 10 - 3.96 = 6.04 g

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If a radioactive nucleus decays according to the following reaction

${}_{72}X_{180} \overset{α}{\to} X_{1} \overset{β}{\to} X_{2} \overset{α}{\to} X_{3} \overset{γ}{\to} X_{4}$

then the mass number and the atomic number of $X_{4}$ then the mass number and the atomic number of $X_{4}$ will be, respectively [MP PET 2002]

172, 70

172, 69

172, 68

171, 69

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Explanation

${}_{Z}P_{A} \to {}_{Z - 2}D_{A - 4}, {}_{Z}P_{A} \to {}_{Z + 1}D_{A} [P s \tan d s f o r p a r e n t D s \tan d s f o r d a u g h t e r n u c l e u s]$

$α$ decay causes decrease in A by 4 and Z decreased by 2 where as in $β$ decay Z increases by 1 and A remains same. In $γ$ decay Z and A remian same.

${}_{72}X_{180} \overset{α}{\to} {}_{70}X_{1}^{176} \overset{β}{\to} {}_{71}X_{2}^{176} \overset{α}{\to} {}_{69}X_{3}^{172} \overset{γ}{\to} {}_{69}X_{4}^{172}$

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Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants 10 $λ$ and $λ$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time

$\frac{1}{10 λ}$

$\frac{1}{11 λ}$

$\frac{11}{10 λ}$

$\frac{1}{9 λ}$

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Explanation

$X_{1} = N_{0} e^{- λ_{1} t} = N_{0} e^{- 10 λ t;} X_{2} = N_{0} e^{- λ_{2} t} = N_{0} e^{- λ t;} \frac{X_{1}}{X_{2}} = \frac{N_{0} e^{- 10 λ t}}{N_{0} e^{- λ t}} = e^{- 9 λ t}; g i v e n \frac{X_{1}}{X_{2}} = \frac{1}{e} ∴ \frac{1}{e} = e^{- 9 λ t} o r e^{- 1} = e^{- 9 λ t}, ∴ 1 = 9 λ t o r t = \frac{1}{9 λ}$

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A radioactive material has mean-lives of 1620 yr and 520 yr for $α$ and $β$ -emission. The material decays by simultaneous $α$ and $β$ -emission. The time in which 1/4th of the material remains intact is

4675 yr

720 yr

545 yr

324 yr

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Explanation

$λ = (λ_{1} + λ_{2}) = \frac{1}{1620} + \frac{1}{520} = 2.54 \times 10^{- 3} Y r^{- 1}$

$∴ t_{1 / 2} = \frac{I n (2)}{λ} = 272.8 y r s .$

$\frac{1}{4}$ th of the material remains intact after 2 half-lives.

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The radioactivity of an element becomes $\frac{1}{64} t h$ of its original value in 60 seconds. Then the half value period is

5 sec

10 sec

20 sec

30 sec

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Explanation

$∵ \frac{1}{64} = {(\frac{1}{2})}^{6}$

$∴$ 6 half life period = 60 sec. or Half life period is 10 sec.

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A nucleus of ${}_{84}^{210}P o$ originally at rest emits $α$ particle with speed $ν$ . What will be the recoil speed of the daughter necleus. [DCE 2002]

4 $ν$ / 206

4 $ν$ / 214

$ν$ / 206

$ν$ / 214

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Explanation

${}_{84}P o^{210} \to {}_{82}X_{206} + {}_{2}H e^{4}$

Using conservation of linear momentum

206 $ν$ '+4 $ν$ = 0 $\Rightarrow$ $ν' = - \frac{4 ν}{(206)} \Rightarrow |ν'| = \frac{4 ν}{206}$

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An atom of mass number 15 and atomic number 7 captures an $α$ -particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be

14 and 2

15 and 3

16 and 4

18 and 8

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Explanation

88. ${}_{7}X_{15} + α - p a r t i c l e \to {}_{9}Y_{19} - {}_{1}H_{1} \to {}_{8}Z_{18}$

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If radius of the ${}_{13}^{27}{Al}$ nucleus is estimated to be 3.6 fermi then the radius of ${}_{52}^{125}T e$ nucleus be nearly

4 Fermi

5 Fermi

6 Fermi

8 Fermi

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Explanation

$\Rightarrow \frac{3.6}{r_{2}} = {(\frac{27}{125})}^{1 / 3} = \frac{3}{5} \Rightarrow r_{2} = 6 f e r m i$

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