Question

# In a photoemissive cell with exciting wavelength λ , the fastest electron has speed v. If the exciting wavelength is changed to 3λ4 , the speed of the fastest emitted electron will be

A
v(34)1/2
B
v(43)1/2
C
Less than v(43)1/2
D
Greater then v(43)1/2

Solution

## The correct option is B Greater then v(43)1/2From E=W0+12mv2max⇒vmax=√2Em−2W0m (where E=hcλ) If wavelength of incident light charges from λ to 3λ4 (decreases) Let energy of incident light charges from E to E’ and speed of fastest electron changes from v to v’ then v=√2Em−2W0m  . . . . . . .(i)  and v′=√2E′m−2W0m   . . . . . .(2) As E∞1λ⇒E′=43E hence v′=√2(43E)m−2W0m⇒v′=(43)12√2Em−2W0m(43)12 ⇒v′=(43)12 √2Em−2W0m(43)12  > v

Suggest corrections