A meacury arec lamp provides `0.1` watt of ultra-violet radiation at a wavelength of `lambda=2537 Å` only. The photo tube (cathode of photo electric device) consists of potessium and has an effective area of `4 cm^(2)`. The cathode is located at a distance of `1 m` from the radiation source. The work funcation for potassium is `phi_(0) = 2.22 eV`.
According to classical theory, the radiation from arc lamp spreads out uniformaly in space as spherical wave. What time of expource to the radiation should be required for a potassium atom (radius `2Å`) ub tge cathode to accumulate sufficient enerfy to eject a photo-electron?

Options

A. `352` second
B. `176` second
C. `704` seconds
D. None time lag

Answer

A

Explanation

UV energy flux at a distance of `1m = (0.1)/(4pi xx 1^(2))`
cross section (effective area) of atom `= pi xx (2 xx 10^-10)^(2) = 4pi xx 10^(20) m^(2)`
Energy required to eject a photoelectron from potassium `= 2.2 eV -= 2.2 xx 1.6 xx 10^(-19) J`.
`rArr "Exposure time" = (2.2 xx 1.6 xx 10^(-19))/(((0.1)/(4pi xx 1^(2))) (4pi xx 10^(-20))) = 352` seconds