JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Modern Physics

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)200✓ Correct
(B)150
(C)400
(D)360

Answer: (A)

Solution

The count rate falls by a factor

1600/100 = 16

over

8

s, and since

16 = 2^{4}

, four half-lives fit into

8

s, giving a half-life of

2

s. At

t = 6

s exactly three half-lives have passed, so the rate is

1600 \times (\frac{1}{2})^{3} = \frac{1600}{8} = 200

counts per second. Recognising

16

as a power of two turns the decay into simple halving and avoids logarithms entirely.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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