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 at t=0 seconds it was 100 count 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

Hint: The number of undecayed radioactive particles is given by the formula $$N=N_0{e}^{-\lambda t}$$. We will use this formula in this question to determine the number of undecayed particles at t=6 seconds. Also we are given two time periods with two disintegrations per second. This will form two equations, so that we can find the number of emitted particles at t=6 seconds.

Formula Used:
$$N=N_0{e}^{-\lambda t}$$
Where:
$$N$$ is number of radioactive particles at time t
$$N_0$$ is number of radioactive particles at t=0 or number of radioactive particles at beginning
$$\lambda$$ is decay constant

Complete step by step answer:
As we know that number of radioactive particles at time t is given by
$$N=N_0{e}^{-\lambda t}$$
Taking log on both sides
$$\begin{array}{rl}& \log N=\log N_0-\lambda t\\ & \lambda t=\log N_0-\log N\end{array}$$
$$\lambda t=\log {\displaystyle \frac{N_0}{N}}$$ …………(i)
At t=0, number of radioactive particles emitting per second are 1600 i.e. $$N_0=1600$$
Now, at t=8 seconds, it is given that number of radioactive elements are 100 i.e. $$N=100$$. Putting this in equation (i) we get,
$$\begin{array}{rl}& \lambda (8)=\log \left({\displaystyle \frac{1600}{100}}\right)\\ & \Rightarrow \lambda (8)=\log 16\\ & \Rightarrow \lambda (8)=\log 2^4\\ & \Rightarrow \lambda (8)=4\log 2\\ & \Rightarrow \lambda ={\displaystyle \frac{\log 2}{2}}\end{array}$$
We have to find the number of radioactive particles emitting at t=6 seconds. Therefore,
$$\lambda t=\log {\displaystyle \frac{N_0}{N}}$$
$$\begin{array}{rl}& {\displaystyle \frac{\log 2}{2}}(6)=\log {\displaystyle \frac{1600}{N}}\\ & \Rightarrow 3\log 2=\log {\displaystyle \frac{1600}{N}}\\ & \Rightarrow \log 2^3=\log {\displaystyle \frac{1600}{N}}\\ & \Rightarrow \log 8=\log {\displaystyle \frac{1600}{N}}\\ & \Rightarrow 8={\displaystyle \frac{1600}{N}}\\ & \Rightarrow N=200\end{array}$$
Hence, the number of radioactive particles emitted per second at t=6 second is 200.
Therefore option C is the correct answer.

Note: We can directly solve this problem by the formula $$N=N_0{e}^{-\lambda t}$$ and just put the given time period one by one. But this makes our solution complicated, so to deal with it easily we just take a long approach. Because in this type of question we are given with such numbers, which are easily cancelled or are factors when taken in fraction form.