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Manipal MET2016PhysicsDual Nature of Matter

In a photoemissive cell with exciting wave length , the fastest electron has a speed v . If the exciting wavelength is changed to 3 / 4 , then the speed of the fastest emitted electron will be

Options

  1. A( 3 4 )^ 1 2
  2. Bv ( 4 3 )^ 1 2
  3. Cless than v ( 4 3 )^ 1 2
  4. Dgreater than v ( 4 3 )^ 1 2

Correct answer

D. greater than v ( 4 3 )^ 1 2

Step-by-step solution

From Einstein's photoelectric equation, 1 2 m v^2= h c -W (i) and 1 2 m v^ 2 = h c 3 / 4 -W (ii) On dividing Eq. (ii) by Eq. (i), we get ( v^ v )^2= 4 h c 3 -W h c -W array ll & W 0 & v^ v ( 4 3 )^ 1 2 array

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