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
\[\begin{align}
  & \log N=\log {{N}_{0}}-\lambda t \\
 & \lambda t=\log {{N}_{0}}-\log N \\
\end{align}\]
\[\lambda t=\log \dfrac{{{N}_{0}}}{N}\] …………(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,
\[\begin{align}
  & \lambda (8)=\log \left( \dfrac{1600}{100} \right) \\
 & \Rightarrow \lambda (8)=\log 16 \\
 & \Rightarrow \lambda (8)=\log {{2}^{4}} \\
 & \Rightarrow \lambda (8)=4\log 2 \\
 & \Rightarrow \lambda =\dfrac{\log 2}{2} \\
\end{align}\]
We have to find the number of radioactive particles emitting at t=6 seconds. Therefore,
\[\lambda t=\log \dfrac{{{N}_{0}}}{N}\]
\[\begin{align}
  & \dfrac{\log 2}{2}(6)=\log \dfrac{1600}{N} \\
 & \Rightarrow 3\log 2=\log \dfrac{1600}{N} \\
 & \Rightarrow \log {{2}^{3}}=\log \dfrac{1600}{N} \\
 & \Rightarrow \log 8=\log \dfrac{1600}{N} \\
 & \Rightarrow 8=\dfrac{1600}{N} \\
 & \Rightarrow N=200 \\
\end{align}\]
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.