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Q.In a sphere of 10^2 m radius, radioactive material emits β^- particles at the rate of 5 x 10^7 s^-1. If 40% of these emitted β^- particles escape from the sphere, how long would it take to raise the potential of the sphere from 0 to 16 V? (Take K = 9 x 10^9 SI unit)?

Gujarat GsebGSEB Higher Secondary Certificate (HSC) Examination 2019Subjective· 3mImportance★★★★★
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Escaping β⁻ particles carry charge off the sample and onto the sphere; the sphere's potential rises as V=kQ/RV=kQ/R, so the time to reach a target potential follows from the rate of charge accumulation.

Note on the given radius: the stem's "10210^2 m" would make the sphere the size of a building, which is not physically sensible for a radioactive sample; this is read as 10−210^{-2} m (1 cm), the standard lab-scale figure for this type of problem, consistent with the other given quantities.

Given emission rate =5×107=5\times10^7 β⁻/s, of which 40%40\% escape =0.4×5×107=2×107=0.4\times5\times10^7=2\times10^7 particles/s escape (each carrying charge e=1.6×10−19e=1.6\times10^{-19} C).

Rate of charge build-up on the sphere:

dQdt=e×(escape rate)=1.6×10−19×2×107=3.2×10−12 C/s\dfrac{dQ}{dt}=e\times(\text{escape rate})=1.6\times10^{-19}\times2\times10^{7}=3.2\times10^{-12}\ \text{C/s}

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