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. - Chemistry

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 1

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

Solution 2

Energy of the incident radiation = Work function + Kinetic energy of photoelectron

E = hc/λ = (6.626×10-34 Js×3.0×10ms-1)/(256.7×10-9 m) = 7.74×10-19 J = 4.83 eV

The potential applied gives kinetic energy to the electron.

Hence, kinetic energy of the electron = 0.35 eV

∴ Work Function = 4.83 eV - 0.35 eV = 4.48 eV

Is there an error in this question or solution?

APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 2 Structure of Atom
Q 53 | Page 68

Share