Question

In question number 5, find the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B.
(Neglect the time taken by the photoelectron to reach plate B)
A. 23 eV
B. 30 eV
C. 15 eV
D. 20 eV

Answer 1

Correct Answer - A
(a) : Charge oon A, `Q_(A)=(5xx10^(7)xx1.6xx10^(-19))`
`=8xx10^(-12)`
Charge on B, `Q_(B)=(33.7-8)xx10^(-12)C=25.7xx10^(-12)C`
`:.E=(sigma_(B))/(2epsilon_(0))-(sigma_(A))/(2epsilon_(0))orE=(1)/(2Aepsilon_(0))(Q_(B)-Q_(A))`
or `E=(1.77xx10^(-12))/(2xx(5xx10^(-4))xx(8.85xx10^(-12)))`
or `E=2000N//C`
Energy of photoelectrons on plate B
energy = E-W=(5-2)eV=3eV
Increase in energy `=(Ed)eV=(2xx10^(3))(10^(-2))eV=20eV`
`:."Energy of photoelectrons on plate"`
B=(20+3)eV=23eV