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General
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Question

Light of wavelength 2475 Å is incident on barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density fraction numerator 1 over denominator square root of 17 end fraction cross times 1 0 to the power of negative 5 end exponentTesla. Work function of the barium is (Given fraction numerator e over denominator m end fraction equals 1.7 cross times 1 0 to the power of 11 end exponent right parenthesis

  1. 1.8 eV    
  2. 2.1 eV    
  3. 4.5 eV    
  4. 3.3 eV    

The correct answer is: 4.5 eV


    Radius of circular path described by a charged particle in a magnetic field is given by r equals fraction numerator square root of 2 m K end root over denominator q B end fraction; where K = Kinetic energy of electron Þ K equals fraction numerator q to the power of 2 end exponent B to the power of 2 end exponent r to the power of 2 end exponent over denominator 2 m end fraction equals open parentheses fraction numerator e over denominator m end fraction close parentheses fraction numerator e B to the power of 2 end exponent r to the power of 2 end exponent over denominator 2 end fraction
    equals fraction numerator 1 over denominator 2 end fraction cross times 1.7 cross times 1 0 to the power of 11 end exponent cross times 1.6 cross times 1 0 to the power of negative 19 end exponent cross times open parentheses fraction numerator 1 over denominator square root of 17 end fraction cross times 1 0 to the power of negative 5 end exponent close parentheses to the power of 2 end exponent cross times left parenthesis 1 right parenthesis to the power of 2 end exponent
    equals 8 cross times 1 0 to the power of negative 20 end exponent J equals 0.5 e V
    By using E = W0 + Kmax
    Þ W subscript 0 end subscript equals E minus K open parentheses fraction numerator 12375 over denominator 2475 end fraction close parentheses subscript m a x end subscript

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