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# A radioactive source in the form of a metal sphere of radius 10-2 m, emits beta particles at the rate of 5 x 1010 partiCles per sec. The source is electrically insulated. How long will it take for it`s potential to be raised by 2 volts, assuming 40% of the emitted beta particles escape the source1. 700 sec 2. 700 µ sec3. 700 milli sec 4. 700 n sec

## 1 Answers

7 years ago

Let t = time for the potential of metal sphere to rise by 2 volt.

Now beta particles escaped in this time = (5x10^10)xt

No. of beta particles escaped in this time = 40% x (5x10^10)x t

= 40/100 x 5 x 10^10 x t

=2 x 10^10 t

So Charge acquired by the sphere in t sec Q = (2 x 10^10 t) x (1.6x10^-19) since we know Q = ne

Q= 3.2 x 10^-9 t coulomb

Radius r = 10^-2 m

The Capacitance C of a metal sphere is given by C = (4 *pie *epsilon )*r

C = [1/(9x10^9)*]*10^-2 = 10^-11/9 farad

We know that Q = C*V (where V = 2 volt)

3.2 x 10^-9 t = 10^-11/9 * 2

t = 700 x 100^-6 sec

Answer (ii) is correct

Thanks & Regards,
Nirmal Singh
Askiitians Faculty

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