An isolated radioactive source in the form a metal sphere of radius $$1$$ mm undergoes β –ray decay at a constant rate of $$6.25$$ × $$1010$$ $$/$$$$sec$$. Assuming $$70$$% of the emitted β –particles escape from the source, time after which potential of the sphere rises by $$1$$ V is

Question

An isolated radioactive source in the form a metal sphere of radius $$1$$ mm undergoes β –ray decay at a constant rate of $$6.25$$ × $$1010$$ $$/$$$$sec$$. Assuming $$70$$% of the emitted β –particles escape from the source, time after which potential of the sphere rises by $$1$$ V is

Infinity Learn · Accepted Answer

$$V=14$$πε$$0qr$$⇒$$1=9$$ x $$109$$ x $$n1e10-3$$ ⇒$$n1=10-39$$ x $$109$$ x $$1.6$$ x $$10-19=10714.4$$ Since $$n1=70100(n)$$ n →no.of β particles emitted ⇒$$n=10070$$ x $$10714.4$$⇒$$n=1087$$ x $$14.4For$$ $$6.25$$ $$x1010$$ particles → $$1secFor$$ $$1087$$ x $$14.4$$ particles →'t' $$sec$$ ∴$$t=10814.4$$ x $$76.25$$ x $$1010=15.5$$ x $$10-6sec=15.5$$ μ $$sec$$

An isolated radioactive source in the form a metal sphere of radius 1 mm undergoes $β$ –ray decay at a constant rate of 6.25 × 10¹⁰ /sec. Assuming 70% of the emitted $β$ –particles escape from the source, time after which potential of the sphere rises by 1 V is

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An isolated radioactive source in the form a metal sphere of radius 1 mm undergoes β –ray decay at a constant rate of 6.25 × 1010 /sec. Assuming 70% of the emitted β –particles escape from the source, time after which potential of the sphere rises by 1 V is

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An isolated radioactive source in the form a metal sphere of radius 1 mm undergoes $β$ –ray decay at a constant rate of 6.25 × 10¹⁰ /sec. Assuming 70% of the emitted $β$ –particles escape from the source, time after which potential of the sphere rises by 1 V is