Ultraviolet light of wavelength 300nn and intensity `1.0Wm^-2` falls on the surface of a photosensitive material. If one per cent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of 1.0 `cm^2` of the surface is nearly

Options

A. `9.61xx10^14s^-1`
B. `4.12xx10^13s^-1`
C. `1.51xx10^12s^-1`
D. `2.13xx10^11s^-1`

Answer

C

Explanation

`n=(power)/((hc)/(lamda))=(300xx10^-9)/(6.6xx10^-34xx3xx10^8)`
`=1.5xx10^18m^-2s^-1=1.5xx10^14cm^-2s^-1`
As only 1 percent of photons cause emission of photo-electrons, number of photo electrons is `n_e=1.5xx10^12s^-1`