A photon of wavelength `6630 Å` is incident on a totally reflecting surface . The momentum delivered by the photon is equal to

Options

A. `6.63 xx 10^(-27) kg - m//sec`
B. `2 xx 10^(-27) kg - m//sec`
C. `10^(-27) kg - m//sec`
D. None of these

Answer

B

Explanation

The momentum of the incident radiation is given as `p = (h)/(lambda)`. When the light is totally reflected normal to the surface the direction of the ray is reversed . That means it reverses the direction of its momentum without changing its magnitude
`:. Delta p = 2 p = ( 2h)/(lambda) = ( 2 xx 6.6 xx 10^(-34))/(6630 xx 10^(-10))`
` = 2 xx 10^(-27) kg - m // sec`.