Question

The half-life of a sample of a radioactive substance is 1 h. If $8\times {10}^{10}$ atoms are present at t= 0, then the number of atoms decayed in the duration t = 2h to t = 4h will be

• A. $2\times {10}^{10}$
• B. $1.5\times {10}^{10}$
• C. zero
• D. Infinity

Answer 1

The correct option is B $1.5\times {10}^{10}$

Suppose $N_0=8\times {10}^{10}atoms$

Counting starts at $t=0$ so $t=2h$ menas $2hour$ later means $\text{two half life later i.e. now the atoms should become 1/(2^2) or 1/4 of initial value}$

Formula used is $N=N_0(1/2{)}^n$ , where $n$ is number of half lives.

So $N_2=N_0/4$

$t=4h$ means $\text{4 half lives from t=0 so the number of atoms should be 1/(2^4) or 1/16 of initial(t=0) value}$

so $N_4=N_0/16$

So number of atoms decayed in time duration between t=2h and t=4h will be $N_2-N_4=N_0(\frac{1}{4}-\frac{1}{16})=3N_0/16$

Putting value of $N_0$ we get the difference as $1.5\times {10}^{10}atoms$

Option B is correct.