CBSE Class 12 Physics 2020 Delhi Set 3 Paper

Question : 3 of 7

Marks: +1, -0

SECTION - C

(a) Define the term 'half-life' of a radioactive substance.
(b) The half life of

{ }_{92}^{238} {\U}

undergoing alpha decay is

4.5 × 10^9

years calculate the activity of

5 {\g}

sample of

{ }_{92}^{238} {\U}

Solution:

(a) Half-life : The half-life of a radioactive substance is the time required for half the number of the nuclei to decay.
(b) Half-life

=4.5 × 10^9

years

\;\text " Mass number "\;\;=238

λ\;=\;{0.693}/{T}=\;{0.693}/{4.5 × 10^9} \;\text " per year "\;

{\N}\;=\;{5 × 6.02 × 10^{23}}/{238}

Hence, the activity

= {\A}=λ {\N}

\;\;\text " Or, "\; \;\; A=\;{0.693}/{4.5 × 10^9} × \;{5 × 6.02 × 10^{23}}/{238}

\; ∴ A=0.00195 × 10^{15}

=1.95 × 10^{12} \;\text " per year "\;

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