The half life of a radioactive substance is T. The time taken, for disintegrating 78^th part of its original mass will be:

Correct answer: (B) 3T. It is given that t_12= T. After disintegration of 78^thpart, 18^th part is only left. Using the decay formula, N=N_012^tt12 ⇒N_08=N_012^tT⇒-ln8=-t/Tln2⇒3ln2=t/Tln2⇒t=3T

Q61 / 90 · JEE Main 2023 (10 Apr Shift 2) · Physics

The half life of a radioactive substance is $T$ . The time taken, for disintegrating ${(\frac{7}{8})}^{t h}$ part of its original mass will be:

A
$2 T$
Correct answer
B
$3 T$
C
$T$
D
$8 T$

Solution

It is given that $t_{\frac{1}{2}} = T$ .
After disintegration of ${(\frac{7}{8})}^{t h}$ part, ${(\frac{1}{8})}^{t h}$ part is only left.
Using the decay formula,
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{t_{\frac{1}{2}}}}$
$\Rightarrow \frac{N_{0}}{8} = N_{0} {(\frac{1}{2})}^{\frac{t}{T}} \Rightarrow - \ln 8 = - \frac{t}{T} \ln 2 \Rightarrow 3 \ln 2 = \frac{t}{T} \ln 2 \Rightarrow t = 3 T$

Hence, the correct answer is option: 'B'.

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The half life of a radioactive substance is T. The time taken, for disintegrating 78th part of its original mass will be:

Solution

Relevant Questions

The half life of a radioactive substance is $T$ . The time taken, for disintegrating ${(\frac{7}{8})}^{t h}$ part of its original mass will be: