A photoelectric surface is illuminated successively by monochromatic light of wavelength `lambda` and `(lambda)/(2)`. If the maximum kinetic energy of the emitted photoelectrons in the second case is `3` times than in the first case , the work function of the surface of the material is
(`h` = Plank's constant , `c` = speed of light )

Options

A. `(hc)/(3lamda)`
B. `(hc)/(2lamda)`
C. `(hc)/(lamda)`
D. `(3hc)/(lamda)`

Answer

B

Explanation

`E=phi` +K.E.
`(hc)/lamda=phi+K.E." "...(i)`
For second case
`(2hc)/lamda=phi+3 K.E.`
`(2hc)/lamda=phi+3((hc)/lamda-phi)" "["From eq. (i)"]`
`(2hc)/lamda=phi+(3hc)/lamda-3phi`
`(2hc)/lamda-(3hc)/lamda=-2phi`
`(-hc)/lamda=-2phi`
`phi=(hc)/(2lamda)`