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CBSE Class 12 · 2005 · All India · Modern PhysicsShort Answer

The half-life of

_{92}^{238} U

against

α

-decay is

4.5 \times 1 0^{9}

years. Calculate the activity of a 1 g sample of

_{92}^{238} U

.

Answer: $A \approx 1.23 \times 1 0^{4} Bq$ .

Solution

Number of nuclei in 1 g:

N = \frac{1}{238} \times 6.023 \times 1 0^{23} = 2.53 \times 1 0^{21} .

Half-life in seconds:

T_{1/2} = 4.5 \times 1 0^{9} \times 3.156 \times 1 0^{7} = 1.42 \times 1 0^{17} s

, so

λ = \frac{0.693}{1.42 \times 1 0 ^{17}} = 4.88 \times 1 0^{- 18} s^{- 1} .

A = λ N = 4.88 \times 1 0^{- 18} \times 2.53 \times 1 0^{21} \approx 1.23 \times 1 0^{4} Bq .

A \approx 1.23 \times 1 0^{4} Bq

CBSE Class 12 Board · 2005 · All India · Subjective

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