SATHEE: Atoms And Nuclei Question 361

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) $6.95 μ s e c$

B) $, 0.95 μ s e c$

C) $1.95 μ s e c,$

D) $2.15 μ s e c$

Show Answer

Answer:

Correct Answer: A

Solution:

Let t = time for the potential of metal sphere to rise by one volt. Now $β -$ particles emitted in this time $= (6.25 \times 10^{11}) \times t$
Number of $β$ -particles escaped in this time $= (80 / 100) \times (6.25 \times 10^{10}) t = 5 \times 10^{10} t$

$∴$ 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) ( $∵$ emission of $β$ -particle lends to a charge eon metal sphere) The capacitance C of a metal sphere is given by $C = 4 π ε_{0} \times r$ $= (\frac{1}{9 \times 10^{9}}) \times (\frac{10^{- 3}}{2}) = \frac{10^{- 12}}{18}$ farad ..(ii) we know that $Q = C \times V$ {Here $V = 1$ volt}

$∴, (8 \times 10^{- 9}) t = (\frac{10^{- 12}}{18}) \times 1$

Solving it for t, we get $t = 6.95 μ s e c .$

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