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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? |
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Answer» 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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