in an experimental set up to study the photoelectric effect a point source of light of power `3.2xx10^(-3)` W was taken. The source can emit mono energetic photons of energy 5eV and is located at a distance of 0.8 m from the center of a stationary metallic sphere of work-function 3.0 eV. The radius of the sphere is `r = 8xx10^(-3)` m. The efficiency of photoelectric emission is one for every `10^6` incident photons. Based on the information given above answer the questions given below. (Assume that the sphere is isolated and photoelectrons are instantly swept away after the emission).
Time after which photoelectric emission stops is

Options

A. `2((KE_(max))/(e))`
B. `((KE_(max))/(e))`
C. `((KE_(max))/(3e))`
D. `((KE_(max))/(2e))`

Answer

B

Explanation

If P is the power of point source of light, the intensity at a distance r is
`I=(P)/(4pir^2)`
The energy intercepted by themetallic sphere is
`E=` intensity `xx` projected area of sphere `=(P)/(4pir^2)xxpiR^2`
If e is the energy of the single photon and `eta` the efficiency of the photon to liberate an electron, the number of ejected electrons is `eta(PR^2)/(4r^2r)=((10^(-6))(3.2xx10^(-3))(8xx10^(-3))^2)/(4xx(0.8)xx(5xx1.6xx10^(-19)))`
`=10^(5)` electron `s^(-1)`
The emission of electrons from a metallic sphere leaves it positively charged. As the potential of the charged sphere begins to rise, it attracts emitted electron. The emission of electrons will stop when the kinetic energy of the electrons if neutralised by the retarding potential of the sphere. So, we have
`eV=(KE_(max))`
`V=((KE_(max))/(e))`
From Einstein's photoelectric equation,
`KE_(max)=hv-phi=(5-3)=2 eV` The potential of a charged sphere is
`V=(1)/(4piepsi_0)(q)/(R)=(1)/(4piepsi_0)((n e)/(R))`
`(1)/(4piepsi_0)((n e)/(R))=2`
`n=(4piepsi_(0)2R)/(e)`
`(2xx8xx10^(-3))/(9xx10^(9)xx1.6xx10^(-19))=1.11xx10^(7)`
The photoelectric effect will stop when `1.11xx10^(7)` electrons have been emitted. The time taken by it to emit `1.11xx10^(7)` electrons,
`t=(1.11xx10^(7))/(10^(5))=111 s=1.85` min