Atoms And Nuclei Question 361
Question: A radioactive source in the form of metal sphere of diameter $ {{10}^{-3}}m $ emits beta particle at a constant rate of $ 6.25\times 10^{10} $ particles per second. If the source is electrically insulated, how long will it take for its potential to rise by 1.0 volt, assuming that 80% of the emitted beta particles escape from the source.
Options:
A) $ 6.95\mu sec $
B) $ ,0.95\mu sec $
C) $ 1.95\mu sec, $
D) $ 2.15\mu sec $
Show Answer
Answer:
Correct Answer: A
Solution:
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Let t = time for the potential of metal sphere to rise by one volt. Now $ \beta - $ particles emitted in this time $ =(6.25\times 10^{11})\times t $
Number of $ \beta $ -particles escaped in this time $ =(80/100)\times (6.25\times 10^{10})t=5\times 10^{10}t $$ \therefore $ Charge acquired by the sphere in t sec. $ Q=(5\times 10^{10}t)\times (1.6\times {{10}^{-19}})=8\times {{10}^{-19}}t $ ..(i) ( $ \because $ emission of $ \beta $ -particle lends to a charge eon metal sphere) The capacitance C of a metal sphere is given by $ C=4\pi {\varepsilon_{0}}\times r $ $ =( \frac{1}{9\times 10^{9}} )\times ( \frac{{{10}^{-3}}}{2} )=\frac{{{10}^{-12}}}{18} $ farad ..(ii) we know that $ Q=C\times V $ {Here $ V=1 $ volt}
$ \therefore ,(8\times {{10}^{-9}})t=( \frac{{{10}^{-12}}}{18} )\times 1 $
Solving it for t, we get $ t=6.95\mu sec. $
BETA
