Manipal MET2014PhysicsDual Nature of Matter
In a photoemissive cell with exciting wavelength , the fastest electron has speed v . If the exciting wavelength is changed by 3 / 4 , the speed of the fastest emitted electron will be
Options
- Av(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 E=W₀+ 1 2 m v^2 v_ = 2 E m - 2 W₀ m (where E= h c ) If wavelength of incident light charges from to 3 4 (decreases). Let energy of incident light charges from E to E^ and speed of fastest electron changes from v to v then aligned v & = 2 E m - 2 W₀ m ...(i) and v & = 2 E^ m - 2 W₀ m ...(ii) aligned and As E 1 E^ = 4 3 E Hence, = 2 ( 4 3 E ) m - 2 W₀ m v^ = ( 4 3 )^ 1 / 2 2 E m - 2 W₀ m ( 4 3 )^ 1 / 2 V= ( 4 3 )^ 1 / 2 X= 2 E m - 2 W₀ m ( 4 3 )^ 1 / 2 v So v^ ( 4 3 )^ 1 / 2 v .