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