Threshold wavelength of a metal is lamda_(0). The de Broglie wavelength of photoelectron when the metal is irradiated with the radiation of wavelength lamda is:

Threshold wavelength of a metal is lamda_(0). The de Broglie wavelengt

1.	Threshold wavelength of a metal is lamda_(0). The de Broglie wavelength of photoelectron when the metal is irradiated with the radiation of wavelength lamda is:
Answer» `[(h lamdalamda_(0))/(2cm)]^((1)/(2))` `[(h(lamda-lamda_(0)))/(2cmlamdalamda_(0))]^(1//2)` `[(h(lamda_(0)-lamda))/(2cmlamdalamda_(0))]^(1//2)` `[(hlamdalamda_(0))/(2mc(lamda_(0)-lamda))]^(1//2)` SOLUTION :Absorbed energy=Threshold energy+Kinetic energy of photoelectron `h(C)/(lamda)=(hc)/(lamda_(0))+E` `E=hc[(1)/(lamda)-(1)/(lamda_(0))]` . . . .(i) DE Broglie wavelength can be CALCULATED as `lamda=(h)/(sqrt(2Em))`.. . . (ii) From eqs. (i) and (ii) `lamda=[(hlamdalamda_(0))/(2mc(lamda_(0)-lamda))]^(1//2)`