1.

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

Answer» Correct Answer - C
`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`


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