Question

The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

NCERT Class XII
Chemistry - Main Course Book I

Chapter 2. Structure of Atom

Answer 1

Work function for Caesium atom = 1.9 eV..
   Threshold frequency = v₀ = f₀ = E / h
          = 1.9 * 1.602 * 10⁻¹⁹ / 6.626 * 10⁻³⁴
          =  4.593 * 10¹⁴ Hz
===========

  Threshold wave length = λ₀ = c / f₀ = 2.98 * 10⁸ / 4.593 * 10¹⁴   meters
              = 648.71 * 10⁻⁹  meters
              = 648.71  nano meters
==========

       f = 500 nm
   Energy in the photons in the irradiation :E = h f 
   Kinetic energy of the emitted electrons KE = h (f - f₀) = h c ( 1/λ - 1/λ₀)

KE = 6.626 * 10⁻³⁴ * 3 * 10⁸ * (1/500 - 1/648.71) * 10⁹   Joules
           =  9.1136 * 10⁻²⁰  Joules
 
mass of electron is  9.1 * 10⁻³¹ kg

=>  1/2 m v² = KE
     =>   v = 4.475 * 10⁵  m/sec
               = 447.5 kilo meters / sec

Answer 2

Answer:

Explanation:

Work function for Caesium atom = 1.9 eV..

Threshold frequency = v₀ = f₀ = E / h

= 1.9 * 1.602 * 10⁻¹⁹ / 6.626 * 10⁻³⁴

= 4.593 * 10¹⁴ Hz

===========

Threshold wave length = λ₀ = c / f₀ = 2.98 * 10⁸ / 4.593 * 10¹⁴ meters

= 648.71 * 10⁻⁹ meters

= 648.71 nano meters

==========

f = 500 nm

Energy in the photons in the irradiation :E = h f

Kinetic energy of the emitted electrons KE = h (f - f₀) = h c ( 1/λ - 1/λ₀)

KE = 6.626 * 10⁻³⁴ * 3 * 10⁸ * (1/500 - 1/648.71) * 10⁹ Joules

= 9.1136 * 10⁻²⁰ Joules

mass of electron is 9.1 * 10⁻³¹ kg

=> 1/2 m v² = KE

=> v = 4.475 * 10⁵ m/sec

= 447.5 kilo meters / sec