Question

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t=0 it was 1600 counts per second and t=8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t=6 seconds is close to

• A. 150
• B. 360
• C. 200
• D. 400

Answer 1

The correct option is C

200

Step 1: Given data

At t=0 it was 1600 counts per second and at t=8 seconds it was 100 counts per second

Step 2: Formula used

Number of nuclei in terms of half-life

$N=\frac{N_0}{2^{\left({\displaystyle \frac{t}{t_{1/2}}}\right)}}$

Here, $N=$quantity remaining, $N_0=$initial quantity, $t=$elapse time and $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=$half-life of the substance.

Step 3: Calculating count rate at t=6

Let A= count rate of emitted particle from radioactive source in count per second.

At $t=0,A_0=1600c/s$

At $t=8,A_8=100c/s$

As we have $N=\frac{N_0}{2^{\left({\displaystyle \frac{t}{t_{1/2}}}\right)}}$

Similarly

$$\begin{gathered} A=\frac{A_0}{2^{\left({\displaystyle \frac{t}{t_{1/2}}}\right)}} \\ {A}_8=\frac{A_0}{2^{\left({\displaystyle \frac{8}{t_{1/2}}}\right)}} \\ \frac{A_0}{A_8}=2\left(\frac{8}{t_{{\displaystyle \frac{1}{2}}}}\right)=\frac{1600}{100}=16=2^4 \\ \frac{8}{t_{{\displaystyle \frac{1}{2}}}}=4 \\ {t}_{\frac{1}{2}}=2sec \\ Sofort=6sec \\ {A}_6=\frac{A_0}{2^{\left({\displaystyle \frac{t}{t_{1/2}}}\right)}}=\frac{A_0}{2^{6/2}}=\frac{A_0}{2^3}=\frac{A_0}{8} \\ {A}_6=\frac{A_0}{8}=\frac{1600}{8}=200c/s \\ {A}_6=200c/s \end{gathered}$$

Hence, option C is the correct option.