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Question: In a photo emissive cell with executing wavelength λ, the fastest electron has speed v. If the excit...

In a photo emissive cell with executing wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be –

A

v (3/4)1/2

B

v (4/3)1/2

C

Less than v (4/3)1/2

D

Greater than v (4/3)1/2

Answer

Greater than v (4/3)1/2

Explanation

Solution

vmax = 3 h m(hcλ−W)\sqrt { \frac { 3 \mathrm {~h} } { \mathrm {~m} } \left( \frac { \mathrm { hc } } { \lambda } - \mathrm { W } \right) }

λ2 < λ1

>

Vmax⁡2\mathrm { V } _ { \max _ { 2 } } =

=

> 43\sqrt { \frac { 4 } { 3 } }