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If a radioactive substance reduces to $\frac{1}{16}$ of its original mass in 40 days, what is its half-life

10 days

20 days

40 days

None of these

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Explanation

(a) $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow \frac{1}{16} = {(\frac{1}{2})}^{4} = {(\frac{1}{2})}^{n} \Rightarrow n = 4$

Also $n = \frac{t}{T_{1 / 2}} \Rightarrow T_{1 / 2} = \frac{40}{4} = 10 days$

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99% of a radioactive element will decay between

6 and 7 half lives

7 and 8 half lives

8 and 9 half lives

9 half lives

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Explanation

(a) $N = N_{0} {(\frac{1}{2})}^{n} \Rightarrow \frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow \frac{1}{100} = {(\frac{1}{2})}^{n} \Rightarrow 2^{n} = 100$

n comes out in between 6 and 7.

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The half-life of a radioactive substance against $α$ -decay is $1.2 \times 10^{7}$ s. What is the decay rate for $4 \times 10^{15}$ atoms of the substance

$4.6 \times 10^{12} atoms / s$

$2.3 \times 10^{11} atoms / s$

$4.6 \times 10^{10} atoms / s$

$2.3 \times 10^{8} atoms / s$

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Explanation

(d) $\frac{dN}{dt} = - λN \Rightarrow |\frac{dN}{dt}| = \frac{0.693}{T_{1 / 2}} \times N$

$= \frac{0.693}{1.2 \times 10^{7}} \times 4 \times 10^{15} = 2.3 \times 10^{8} atoms / \sec$

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In a sample of radioactive material, what percentage of the initial number of active nuclei will decay during one mean life

69.3%

63%

50%

37%

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Explanation

(b) Number of atoms remains undecayed $N = N_{0} e^{- λt}$

Number of atoms decayed = $N_{0} (1 - e^{- λt})$

= $N_{0} (1 - e^{- λ \times \frac{1}{λ}}) = N_{0} (1 - \frac{1}{e}) = 0.63 N_{0} = 63 % of N_{0}$

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The S.I. unit of radioactivity is

Roentgen

Rutherford

Curie

Becqueral

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Explanation

(d) The S.I. unit of radioactivity is Becqueral

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A radioactive material has an initial amount 16 gm. After 120 days it reduces to 1 gm, then the half-life of radioactive material is

60 days

30 days

40 days

240 days

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Explanation

(b) $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow (\frac{1}{16}) = {(\frac{1}{2})}^{n} \Rightarrow n = 4$

also $n = \frac{t}{T_{1 / 2}} \Rightarrow T_{1 / 2} = \frac{120}{4} = 30 days$

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The equation ${}_{Z}^{A}X \to {}_{Z + 1}^{A}Y + {}_{- 1}^{0}e + \bar{v}$ represents-

$β$ -emission

$α$ -emission

$e^{-}$ capture

Fission

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Explanation

(a) It is a $β - d e c a y$ as the atomic number(Z) increases by one.

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The activity of a sample of a radioactive material is $A_{1}$ , at time $t_{1}$ and $A_{2}$ at time $t_{2}$ ( $t_{2}$ > $t_{1}$ ). If its mean life T, then

$A_{1} t_{1} = A_{2} t_{2}$

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

$A_{2} = A_{1} e^{(t_{1} - t_{2}) / T}$

$A_{2} = A_{1} e^{(t_{1} / t_{2}) T}$

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Explanation

So $A_{1} = A_{0} e^{- t_{1} / T} \Rightarrow A_{0} = \frac{A_{1}}{e^{- t_{1} / T}} = A_{1} e^{t_{1} / T}$

$∴ A_{2} = A_{0} e^{- t / T} = (A_{1} e^{t_{1} / T}) e^{- t_{2} / T} \Rightarrow A_{2} = A_{1} e^{(t_{1} - t_{2}) / T}$

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Nucleus produced due to $α$ -decay of the nucleus ${}_{Z}^{A}X$ is

${}_{Z + 2}^{A + 4}Y$

${}_{Z}^{A}Y$

${}_{Z - 2}^{A - 4}Y$

${}_{Z - 4}^{A - 2}Y$

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Explanation

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The half-life of ${}_{215}{At}$ is 100 $μs$ . The time taken for the radioactivity of a sample of ${}_{215}{At}$ to decay to 1/16th of its initial value is

400 $μs$

6.3 $μs$

40 $μs$

300 $μs$

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Explanation

(a) The radioactivity of a sample decays to $\frac{1}{16} th$ of its initial value in four half lives.

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