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A radioavtive source in the form of a me...

A radioavtive source in the form of a metal sphere of daimeter `10^(-3)` m emits `beta`-particles at a constant rate of `6.25 xx 10^(10)` particles per second. If the source is electrically insulated, how long will it take for its potential to rise by `1.0 V`, assuming that `80%` of the emitted `beta`-particles escape the socurce?

Text Solution

Verified by Experts

As only `80%` of emitted `beta` particles escape form the source, therefore, Number of `beta` particles leaveing the source/sec
`=80/100xx6.25xx10^(10)=5xx10^(10)`
Positive charge gained/sec by the source
`q=5xx10^(10)xx(1.6xx10^(-19))C=8xx10^(-9)C`
Capacity of sphere, `C=4pi in_0r`
`=1/(9xx10^9)((10^(-3))/2)=(10^(-12))/18F`
Increase of potential of sphere per sec
`=("Increase of charge"//"sec")/("Capacity")`
`=8xx10^(-9)xx18/(10^(-12))=1.44xx10^5V`
Time taken for rise of potential by one volt.
`=1/(1.44xx10^5)=6.94xx10^(-6)s`
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