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13.Nuclei
hard
The rate of disintegration was observed to be ${10^{17}}$ disintegrations per sec when its half life period is $1445$ years. The original number of particles are
A
$8.9 \times {10^{27}}$
B
$6.6 \times {10^{27}}$
C
$1.4 \times {10^{16}}$
D
$1.2 \times {10^{17}}$
Solution
(b) Rate of disintegration $\frac{{dN}}{{dt}} = {10^{17}}{s^{ – 1}}$
Half life ${T_{1/2}} = 1445\, year$
$= 1445 × 365 ×24 × 60 ×60 = 4.55 ×10^{10} \, sec$
Now decay constant
$\lambda = \frac{{0.693}}{{{T_{1/2}}}} = \frac{{0.693}}{{4.55 \times {{10}^{10}}}} = 1.5 \times {10^{ – 11}}\,per \,sec$
The rate of disintegration
$\frac{{dN}}{{dt}} = \lambda \times {N_0} \Rightarrow {10^{17}} = 1.5 \times {10^{ – 11}} \times {N_0}$
==> $N_0 = 6.6 ×10^{27}$
Standard 12
Physics
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