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KCET2015PhysicsRay Optics

Find the de-Broglie wavelength of an electron with kinetic energy of ( 120 eV ).

Options

  1. A( 95 pm )
  2. B( 102 ) pm
  3. C( 112 pm )
  4. D( 124 pm )

Correct answer

C. ( 112 pm )

Step-by-step solution

Given, kinetic energy of electron =120 eV de Broglie wavelength is related to V as = 1.227 ~nm V = 1.227 10⁻⁹ 120 = 1.227 10⁻⁹ 10.95 =0.112 10⁻⁹ Therefore, =112 10⁻¹² ~m =112 pm

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