# 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. - Chemistry

Numerical

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.

#### Solution

From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,

E = W0 + K.E

⇒ W0 = E – K.E

Energy of incident photon (E) = ("hc")/lambda

Where,

h = Planck’s constant

Substituting the values in the given expression of E:

"E" =((6.626 xx 10^(-34) " Js")(3.0xx10^8 " ms"^(-1)))/(256.7xx10^(-9) " m")

= 7.744 xx 10^(-19) " J"

= (7.744xx10^(-19))/(1.602xx10^(-19)) " ev"

E = 4.83 eV

The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,

K.E = 0.35 V

K.E = 0.35 eV

∴Work function, W0 = E – K.E

= 4.83 eV – 0.35 eV

= 4.48 eV

(5lambda_0 - 2000)/(4lambda_0 - 2000) = (5.35/2.55)^2 = 28.6225/6.5025

(5lambda_0 - 2000)/(4lambda_0 - 2000) = 4.40177

17.6070lambda_0 - 5lambda_0= 8803.537 - 2000

lambda_0 = (6805.537)/(12.607)

lambda_0 = 539.8 " nm"

lambda_0 = 540 " nm"

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Chapter 2: Structure of Atom - EXERCISES [Page 72]

#### APPEARS IN

NCERT Chemistry Part 1 and 2 Class 11
Chapter 2 Structure of Atom
EXERCISES | Q 2.53 | Page 72

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