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The work function of a metal is2,5 xx (10^-19) J. (a) Find the threshold frequency for photoelectric emission. (b) If the metal is exposed to a light beam of frequency (6.0 xx 10^14) Hz, what will be the stopping potential ? |
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Answer» Solution :GIVEN ` W_0 = 2.5 XX 10^(-19) J ` (a) we have `W_0 = hv_0` or ` V_0 = W_0 / H ` `=2.5 xx 10^(-19) / 6.063 xx10^(-34) ` `= 3.77 xx 10^(14) Hz` `=3.8 xx 10^(14) Hz` (B)` eV_0 = hv - W_0 ` or, V_0 = hv - W_0 / E ` ` = 6.63 xx 10^(-34) xx 6 xx 10^(14) - 2.5 xx 10^(-19) / 1.6 xx 10^(-19) ` `= 6.97 xx 10^(-19) - 2.5 xx 10^(-19) / 1.6 xx 10^(19) ` `0.91 V` |
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