solve 11) Q 11 Light of wavelength 300 nm is incident on a photosensitive surface The stopping potential - Physics -

Question

solve 11)
Q 11 Light of wavelength 300 nm is incident on a photosensitive
surface The stopping potential for emitted photoelectrons is 2 5 v
The wavelength of incident light is reduced to 150 nm The stopping
potential of emitted photoelectron is

Sanjay Singh · Accepted Answer

Dear student
Using the formula e×Vs=hcλ-W (W:work function)e×2
5=hc300(nm)-We×2 5=1241 5 eV×nm300(nm)-We×2
5=1241 5 eV300-We×2 5=1241 5 ×(1
6×10-19)joule300-We×2 5=1241
5 ×e (joule)300-W 1e×Vs=1241
5 ×e (joule)150-W
2subtracting eqn 1 from 2e×Vs-e×2
5=1241 5 ×e (joule)150-1241
5 ×e (joule)300(Vs-2 5)=1241 5 150-1241
5300(Vs-2 5)=1241 5 150-1241 5300=1241 5 300=4
138 vVs=2 5+4 138=6 63 voltRegards

solve 11) Q.11. Light of wavelength 300 nm is incident on a photosensitive surface. The stopping potential for emitted photoelectrons is 2.5 v. The wavelength of incident light is reduced to 150 nm. The stopping potential of emitted photoelectron is

solve 11)
Q.11. Light of wavelength 300 nm is incident on a photosensitive surface. The stopping potential for emitted photoelectrons is 2.5 v. The wavelength of incident light is reduced to 150 nm. The stopping potential of emitted photoelectron is