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The difference between the incident energy and threshold energy for an electron in a photoelectric effect experiment is `5 eV`. The de Broglie wavelength of the electron is-A. `(6.6 xx 10^-9)/(sqrt(1456)) m`B. `(6.6 xx 10^-9)/(sqrt(145.6)) m`C. `(6.6 xx 10^-9)/(sqrt(1664))`D. `(6.6 xx 10^-9)/(sqrt(166.4))m` |
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Answer» Correct Answer - B (b) `KE of overline (e)^(-)` ejected = Energy of incident quantum - threshold energy `= 5 eV` `lamda = (h)/(mV) = (h)/(sqrt(2 m K.E))` =`(6.6 xx 10^-34)/(sqrt(2 xx 9.1 xx 10^-31 xx 5 xx 1.6 xx 10^-19))` `m = (6.6)/(sqrt(145.6)) m`. |
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