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A metal plate of area 1 xx 10^(-4) m^(2)...

A metal plate of area `1 xx 10^(-4) m^(2)` is illuminated by a radiation of intensity `16m W//m^(2)`. The work function of the metal is 5eV. The energy of the incident photons is 10eV. The energy of the incident photons is 10eV and 10% of it produces photo electrons. The number of emitted photo electrons per second their maximum energy, respectively, will be : `[1 eV = 1.6 xx 10^(-19)J]`

A

`10^(10) and 5eV`

B

`10^(11) and 5eV`

C

`10^(12) and 5eV`

D

`10^(14) and 10 eV`

Text Solution

Verified by Experts

The correct Answer is:
B
Maximum kinetic energy `KE_("max") = E - phi`
`KE_("max") = 10 eV - 5eV = 5eV `
No . of photons incident per unit time `n/t = (IA)/E`
`n/t = (16xx10^(-3)xx10^(-4))/(10 xx 1.6 xx 10^(-19))= 10^(12)`
No of photoelectrons ejected per unit time
10 % of `n/t = 10/100 xx 10^(12) = 10^(11)`
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