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In a photoelectric experiment, it was found that the stopping potential decreases from 1.85 V to 0.82 V as the wavelength of the incident light is varied from 300nm to 400nm. Calculate the value of the Planck constant from these data. |
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Answer» Solution : The maximum kinetic energy of a photoelectron is ` (K_max = hc/lambda - varphi)` and the stopping POTENTIAL is ` V= (K_max/e)= (hc/lambda e )- (varphi/ e).` ` If V_1, V_2, are the stopping potentials at wavelengths lambda_1 and lambda_2 respectively, ` V_1 = (hc/lambda_1 e)- (varphi/e)` ` and V_2= (hc/ (lambda_2 e)-(varphi/ e)).` ` This gives, V_1- V_2 = (hc/e) ( 1/lambda_1) - (1 / lambda_2) ` or, h= (e((V_1)-(V_2))/ C(1/lambda_1)-(1/lambda_2))` ` = (e(1.85 V - 0.82V)/c(1/(300 XX (10^-9)m))-(1/400 xx (10^-9)m))` ` = (1.03e V/ ((3 xx (10^8)m (s^-1))(1/12 xx (10^7)(m^-1)))` ` = 4.12 xx (10^-15) EVS.` |
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