A radioactive element has a rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

Question

A radioactive element has a rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is (A) 0.2loge 2  (B)  0.5loge 2  (C) 0.6loge 2  (D) 0.8loge 2

Tardigrade Admin · Accepted Answer

(N/N0)=e- λ t ∴ (2500/10000)=e- λ × 4 ∴ (1/4)=e- 4 λ ∴ e4 λ =4 ∴ 4λ =loge 4 ∴ 4λ =loge 22 ∴ 4λ =2loge 2 ∴ λ =(2/4)loge 2 ∴ λ =0.5loge 2

Q. A radioactive element has a rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

$0.2 l o g_{e} 2$

$0.5 l o g_{e} 2$

$0.6 l o g_{e} 2$

$0.8 l o g_{e} 2$

$\frac{N}{N _{0}} = e^{- λ t}$
$∴ \frac{2500}{10000} = e^{- λ \times 4}$
$∴ \frac{1}{4} = e^{- 4 λ}$
$∴ e^{4 λ} = 4$
$∴ 4 λ = l o g_{e} 4$
$∴ 4 λ = l o g_{e} 2^{2}$
$∴ 4 λ = 2 l o g_{e} 2$
$∴ λ = \frac{2}{4} l o g_{e} 2$
$∴ λ = 0.5 l o g_{e} 2$

Q. A radioactive element has a rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

0.2loge​2

0.5loge​2

0.6loge​2

0.8loge​2

N0​N​=e−λt ∴100002500​=e−λ×4 ∴41​=e−4λ ∴e4λ=4 ∴4λ=loge​4 ∴4λ=loge​22 ∴4λ=2loge​2 ∴λ=42​loge​2 ∴λ=0.5loge​2

$0.2 l o g_{e} 2$

$0.5 l o g_{e} 2$

$0.6 l o g_{e} 2$

$0.8 l o g_{e} 2$

$\frac{N}{N _{0}} = e^{- λ t}$
$∴ \frac{2500}{10000} = e^{- λ \times 4}$
$∴ \frac{1}{4} = e^{- 4 λ}$
$∴ e^{4 λ} = 4$
$∴ 4 λ = l o g_{e} 4$
$∴ 4 λ = l o g_{e} 2^{2}$
$∴ 4 λ = 2 l o g_{e} 2$
$∴ λ = \frac{2}{4} l o g_{e} 2$
$∴ λ = 0.5 l o g_{e} 2$