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A radiation of wavelength 300 nm is incident on a silver surface. Will photoelectrons be observed ? |
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Answer» Solution :Energy of the incidet photon is `E = h upsilon = (hc)/(lambda)` (in JOULES) `E = (hc)/(lambda e)` (in eV) Substituting the known VALUES, we get `E = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(300 xx 10^(-9) xx 1.6 xx 10^(-19))` E = 4.14 eV the WORK function of SILVER = 4.7 eV. Since the energy of the incident photon is less than the work function of silver, photoelectrons are not observed in this case. |
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