First slide

Photo Electric Effect

Question

Ultra violet light of wavelength 66.26 nm and intensity $2 w / m^{2}$ falls on potassium surface by which photoelectrons are ejected out. If only 0.1% of the incident photons produce photoelectrons and surface area of metal surface is $4 m^{2}$ , how many electrons are emitted per second ?

Moderate

A

$2 .67 \times 10^{15}$
B

$3 \times 10^{15}$
C

$3 .33 \times 10^{17}$
D

$4 .17 \times 10^{16}$

Solution

Number of photons falling on metal surface
$Energy per quanta = \frac{hc}{λ} = \frac{12400}{66 .26 \times 10} eV = 1 .87 \times 10^{1} \times 1 .6 \times 10^{- 19} Joule$
$= 3 \times 10^{- 18} Joule n_{p} = \frac{Intensity \times Area}{Energy per quanta} = \frac{2 \times 4}{3 \times 10^{- 18}} = 2 .67 \times 10^{18} / s$
$From the question, \Rightarrow n_{e} = 0 .1 % of n_{p} = \frac{0 .1}{100} \times 2 .67 \times 10^{18} / s = 2 .67 \times 10^{15} / s .$

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