In a photoelectric experiment a parallel beam of monochromatic light with power of 200 is incident on a perfectly absorbing cathode of work function 6.25. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force 10 due to the impact of the electrons. The value of is __________. Mass of the electron 910 and 1.01.610.

Answer

[24.00]

Explanation

Ejection of one electron requires energy of 6.25 eV, hence number of electrons emitted per second due to 200 W beam can be calculated as
`N=200/(6.25xx1.6xx10^(-19))`
Kinetic energy of electron before hitting the target can be calculated by multiplying charge on electron with potential difference applied. If K is the Kinetic energy of electron, then linear momentum of one electron can be written as `p = sqrt(2mK)`. As electrons are absorbed, by multiplying momentum of electron with number of electrons per second, we can calculate rate of change of linear momentum which is equal to force applied on the target.
`F=Nsqrt(2mK)`
`F = sqrt ((200xx200xx2xx9xx10^(-31)xx500)/(6.25xx6.25xx1.6xx10^(-19)))`
`F=200/(1.6xx6.25xx10^(-19))xxsqrt(144xx10^(-48))`
` = 24.0 xx 10 ^(-4) N`
Hence on comparing we get n = 24.