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. `(2hc)/(lambda)`
B. `(hc)/(3lambda)`
C. `(hc)/(2lambda)`
D. `(hc)/(lambda)`

Answer

3

Explanation

According to Einstein.s photoelectric equation, the maximum kinetic energy of the emitted photoelectric in the first case is
`K_(max1)=(hc)/(lambda)-phi_(0)`
And that in the second case in
`K_(max2) = (hc)/((lambda)/(2))-phi_(0)=(2hc)/(lambda)-phi_(0)`
But ` K_(max2) = 3K_(max1)` (given)
`phi_(0) = (hc)/(2lambda)`.