An EM wave of waalength `lambda` is incident on a photosensitive surface of negligible work function. If m mass is of photo electron emitted from the surface has de-broglie wavelength `lambda_d` then,

Options

A. `lamda_d=((2mc)/h)lamda^2`
B. `lamda=((2mc)/h)lamda_d^2`
C. `lamda=((2h)/(mc))lamda_d^2`
D. `lamda=((2m)/(hc))lamda_d^2`

Answer

B

Explanation

`(hc)/(lamda)=1/2 mv^2 +phi`
`(hc)/lamda=1/2mv^2 " "` ( `:.` given `phi ~~0` )
`(hc)/lamda=(p^2)/(2m)`
`P=sqrt((2mhc)/(lamda))`
`lamda_d = h/P=h/(sqrt((2mhc)/lamda))=sqrt((hlamda)/(2mc))`
`lamda_d^2=(nlamda)/(2mc)implies lamda =(2mc)/hlamda_d^2`