Question

The activity of a radioactive sample is measured as \(N_0\) counts per minute at time \(t = 0\) and \(\frac{N_0}{e}\) counts per minute at time \(t = 3\) minute. The time (in minute) in which the activity reduces to half the value, is

• A. \(\frac{1}{3}log_e2\)
• B. \(3log_e2\)
• C. \(3log_{10}2\)
• D. \(\frac{3}{log_e2}\)

Hint:

The radius of the nth Bohr orbit is proportional to n squared.

Answer 1

The Correct Option is B

Step 1: Bohr's quantisation:
Angular momentum $mvr = \frac{nh}{2\pi}$, with the force balance $\frac{mv^2}{r} = \frac{ke^2}{r^2}$.

Step 2: Solve:
Eliminating v gives $r = \frac{n^2h^2}{4\pi^2mke^2}$, proportional to $n^2$. So the ratio is $1:4:9:16$ (C).

Final Answer:
$1:4:9:16$. \[ \boxed{1 : 4 : 9 : 16} \]