Light of intensity `10^(-5)Wm^(-2)` falls on a sodium photocell of surface area `2cm^(2)`. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave picture of radiation. The work function of the metal is given to be about 2eV. What is the implication of your answer? effective atomic area `=10^(-20)m^(2)`.

Options

A. `10(1)/(2)s`
B. `(1)/(2)s`
C. `(1)/(2)h`
D. `(1)/(2)yr`

Answer

D

Explanation

Incident power of light
`=n'=nxx(A)/(A_(e))`
`=5xx(2xx10^(-4))/(10^(-20))=10^(17)`
Where `A_(e)=`effective area of sodium atom energy absorbed per second by each electron
`E=P//n'=2xx10^(-9)//10^(-17)JS^(-1)`
`therefore`Time required for protoemission in wave picture of light
`=phi//E=(2xx1.6xx10^(-19)J)/(2xx10^(-26))=16xx10^(6)s=0.507yr`