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

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2015-05-30T19:46:44+05:30

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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

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