JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Modern Physics

A freshly prepared sample of a radioisotope of half-life

1386 s

has activity

1 0^{3}

disintegrations per second. Given that

ln 2 = 0.693

, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first

80 s

after preparation of the sample is

Answer: 4

Solution

Decay constant:

λ = \frac{l n 2}{T _{1/2}} = \frac{0.693}{1386} = \frac{1}{2000} s^{- 1}

Initial number of nuclei:

N_{0} = \frac{A _{0}}{λ} = \frac{1 0 ^{3}}{1/2000} = 2 \times 1 0^{6}

Number remaining after time

t

N = N_{0} e^{- λ t}

Number decayed:

N_{0} - N = N_{0} (1 - e^{- λ t})

Percentage decayed:

% = (1 - e^{- λ t}) \times 100 = (1 - e^{- 80/2000}) \times 100

= (1 - e^{- 0.04}) \times 100

Using

e^{- 0.04} \approx 0.961

% \approx (1 - 0.961) \times 100 = 3.9% \approx 4%

The answer is 4.

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Integer answer