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The maximum kinetic energy of the photoelectrons emitted is doubled when the wavelength of light incident on the photosensitive surface changes from lambda_(1)to lambda_(2). Deduce expressions for the threshold wavelength and work function for the metal surface in terms of lambda_(1) and lambda_(2). |
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Answer» Solution :In terms of WAVELENGTH `LAMBDA`, the photoelectric equation may be expressed as: `HC((1)/(lambda)-(1)/(lambda_(0)))=K_("max")`, where `lambda_(0)` is the THRESHOLD wavelength. As per question for incident light of wavelength`lambda_(1), K_("max")=`K(say) and for light of wavelength `lambda_(2), K_("max")=2K`. Hence, we have `hc((1)/(lambda_(1))-(1)/(lambda_(0)))=K and hc((1)/(lambda_(2))-(1)/(lambda_(0)))=2K` `rArr 2((1)/(lambda_(1))-(1)/(lambda_(0)))=(1)/(lambda_(2))-(1)/(lambda_(0)) rArr (1)/(lambda_(0)) =(2)/(lambda_(1))-(1)/(lambda_(2)) rArr lambda_(0)=(lambda_(1)lambda_(2))/((2lambda_(2)-lambda_(1)))` and work function of the metal `phi_(0)=(hc)/(lambda_(0))=(hc(2lambda_(2)-lambda_(1)))/(lambda_(1)lambda_(2))` |
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