JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Modern Physics

To determine the half life of a radioactive element, a student plots a graph of

ℓ n \frac{dN ( t )}{dt}

versus

t

. Here

\frac{d N ( t )}{d t}

is the rate of radioactive decay at time

t

. If the number of radioactive nuclei of this element decreases by a factor of

p

after 4.16 years, the value of

p

Answer: 8

Solution

The rate of decay is:

\frac{d N}{d t} = - λ N = - λ N_{0} e^{- λ t}

Taking natural log:

ln \frac{d N}{d t} = ln (λ N_{0}) - λ t

From the graph, the slope is

- λ = - \frac{1}{2}

per year.

Therefore,

λ = 0.5

per year.

The half-life is:

t_{1/2} = \frac{0.693}{λ} = \frac{0.693}{0.5} = 1.386 years

Time elapsed:

4.16

years =

3 \times 1.386

years =

3 t_{1/2}

After 3 half-lives, the number decreases by a factor of

2^{3} = 8

.

The answer is 8.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer

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