A metal plate of area $1 \times 10^{- 4} m^{2}$ is illuminated by a radiation of intensity $16 \frac{mW}{m^{2}}$ . The work function of the metal is $5 eV$ . The energy of the incident photons is $10 eV$ and only $10 %$ of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be $[1 eV = 1.6 \times 10^{- 19} J]$

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$10^{10}$ and $5 e V$

$10^{11}$ and $5 e V$

$10^{14}$ and $5 e V$

$10^{12}$ and $5 e V$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$ϕ = 5 eV, E = 10 eV ∴ {KE}_{\max} = E - ϕ = 10 - 5 = 5 eV$

Energy incident on the plate,

$E' = (16 \times 10^{- 3}) \times (1 \times 10^{- 4}) \frac{J}{s} = 16 \times 10^{- 7} \frac{J}{s} = 10^{13} \frac{eV}{s}$

Number of photo electrons emitted per second

$= 10 %$ of $\frac{E'}{E} = 0.1 \times \frac{10^{13} \frac{e V}{s}}{10 e V} = 10^{11} s^{- 1}$

Watch 3-min video & get full concept clarity