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If the threshold wavelength `(lambda_(0))` for spection of electron from metal is `350nm `then work function for the photoelectric emission isA. `1.2 xx 10^(-18) J`B. `1.2 xx 10^(-20) J`C. `6 xx 10^(-29) J`D. `6 xx 10^(-12) J` |
Answer» Correct Answer - B Work function = Threshold energy `= hv_(0) = (hc)/(lambda_(0))` ` = (6.6 xx 10^(-34) J s xx 3 xx 10^(8) m)/(330 xx 10^(-9) m) = 6.6 xx 10^(-29) J` |
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