If an em wave of wavelength `lambda` is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength `lambda_1`, prove that
`lambda=((2mc)/h)lambda_1^2`

Options

A. `lamda=(mc)/(h)lamda^('2)`
B. `lamda=(3mc)/(2h)lamda^('2)`
C. `lamda=(2mc)/(h)lamda^('2)`
D. `lamda=(5mc)/(h)lamda^('2)`

Answer

C

Explanation

(c ) : Kinetic energy of emitted electron = Energy of incident photon
`i.e.(1)/(2)mv^(2)=hv`
or `(p^(2))/(2m)=(hc)/(lamda)" "( :.mv=p,v=(c)/(lamda))`
or `p=sqrt((2mhc)/(lamda))`
de Broglie wavelength of emitted electrons
`lamda^(')=(h)/(p)=(h)/sqrt((2mhc)/(lamda))orlamda^(')=sqrt((hlamda)/(2mc)):.lamda=(2mc)/(h)lamda^('2)`