In a photoelectric experiment a metal plate of work function $ϕ = 2.0 e V$ is irradiated with beam of monochromatic light. The photoelectric current (i) versus the applied potential difference (V) graph is as shown in the figure It is known that the efficiency of photoemission is $10^{- 3} % .$ Calculate the power of light incident on the metal plate (in watt)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 7.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Stopping potential difference = 5 volt
$∴$ Energy of incident photon $h υ = 5 + ϕ = 7 e V$
Saturation current is $i_{s} = 10 μ A$
$∴$ Number of electrons emitted per second is $n_{e} = \frac{i_{s}}{e} = \frac{10 \times 10^{- 6}}{1.6 \times 10^{- 19}} = \frac{1}{1.6} \times 10^{14}$
No of photons incident per second
$∴$ Power incident $= (7 e V) n_{p} = (7 \times 1.6 \times 10^{- 19} J) \times (\frac{10^{19}}{1.6} s^{- 1}) = 7 J s^{- 1} = 7 W$