The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Asked by Abhisek | 1 year ago |  68

#### 1 Answer

##### Solution :-

As per the law of conservation of energy, the energy associated with an incident photon (E) must be equal to the sum of its kinetic energy and the work function (W0) of the radiation.

E = W0 + K.E

⇒ W0 = E – K.E

Energy of incident photon

$$(E)= \dfrac{hc}{\lambda }$$

Where,

c denotes the velocity of the radiation

h is Planck’s constant

λ is the wavelength of the radiation

Substituting these values in the expression for E:

$$E=\dfrac{(6.626\times 10^{-34}Js)(3\times 10^{8}ms^{-1})}{(256.7\times 10^{-9}m)}$$

$$=7.744\times 10^{-19}\, J\\ \\$$

$$=\dfrac{7.744\times 10^{-19}}{1.602\times 10^{-19}}eV\\ \\$$

$$E=4.83\, eV$$

The potential that is applied to the silver is transformed into the kinetic energy (K.E) of the photoelectron.

Hence,

K.E = 0.35 V

K.E = 0.35 eV

Therefore, Work function, W0 = E – K.E

= 4.83 eV – 0.35 eV

= 4.48 eV

Answered by Pragya Singh | 1 year ago

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