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Maximum velocity of photoelectron emitted is `4.8 ms^(-1)`. If e/m ratio of electron is `1.76xx10^(11)Ckg^(-1)`, then stopping potential is given byA. `(v^(2))/(2(m/e))`B. `(v^(2))/(2(e/m))`C. `(v^(2))/((e/m))`D. `(v^(2))/((m/e))` |
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Answer» Correct Answer - B Given maximum velcity of photoelectron = v charge of phoroelectron = e and mass of photoelectron = m Let the stopping potential of the photoelectron = v, Then, the maximum kinetic energy ` KE_ ( max ) = eVrArr ( 1 ) / ( 2 ) mv ^ 2 = eV ( because KE= ( 1 ) / ( 2 ) mv ^ 2 ) ` `rArr V = ( v^ 2 m ) /( 2 e ) or V = ( v ^ 2 ) / ( 2 ((e ) / ( m ) ) ) ` |
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