(a) Photoelectric threshold of metallic silver is `lambda =3800 Å`. Ultravilolet light of `lambda =260` nm is incident on silver surface. Calculate :
(i) the value of work function in joule and eV
`K_(max)` of the emitted photoelectrons
(iii) `v_(max)` of the photoelectrons
(b) Ultraviolet light of wavelength 280 nm is used in an experiment on photoelectric effict with lithium `(phi =2.5 eV)` cathode. Find
(i) the maximum kinetic energy of the photoelectrons
(ii) the stopping potential
(c) Find the maximum magnitude of the linear momentum of a photelectron emitted when light of wavelength 400 nm falls on a metal `(phi =2.5 eV)`.
(d) A monochromitc light source of intensity 5 mW emits `8xx10^(5)` photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this set up is `2.0 V` . Calculate the work function of the metal.
(e) The maximum kinetic energy of photoelectrons emitted from a certain metallic surface is 30 eV when imonochromatic radiation of wavelength `lambda` falls on it. When the same surface is illuminated with light of wavelength `2 lambda` the maximum kinetic energy of photo electrons is observed to be 10 eV. Calculate the wavelength `lambda` and determine the maximum wavelength of incident radiation for which photoelectrons can be emitted by this surface.

Explanation

(a) `lambda_(0) =3800 Å =380 nm, lambda =260 nm`
(i) `phi =(1242)/(lambda_(0)(nm))eV =(1242)/(380) =3.3 eV`
`=3.3xx16xx10^(-19) =5.3xx10^(-19) J`
(ii) `E =(1242)/(lambda(nm))eV =(1242)/(260) =4.8 eV`
`E=phi + K_(max) implies 4.8 =3.3 + K_(max)`
`K_(max) =1.5 eV`
(iii) `K_(max) =1.5 eV =1.5xx1.6xx10^(-19) J`
`K_(max) =(1)/(2)mv_(max)^(2)`
`v_(max) =sqrt((2K_(max))/(m)) =sqrt((2xx1.5xx1.6xx10^(19))/(9.1xx10^(-31)))`
`=0.73xx10^(6) m//sec`
(b) `E =(1242)/(lambda(nm))eV =(1242)/(280) =4.48 eV`
`E = phi + K_(max)` `implies 4.4 =2.5 + K_(max)`
`K_(max) =1.9 eV`
`K_(max) =eV_(s) implies V_(s) =1.9 V`
(c) `E=(1242)/(400) =3.1 eV, phi =2.5 eV`
`E =phi + K_(max) implies 3.1 =2.5 + K_(max)`
`K_(max) =0.6 eV`
`P= sqrt(2mK_(max)) =sqrt(2xx9.1xx10^(-31)xx0.6xx1.6xx10^(-19))`
`=4.2xx10^(-25) kg m//sec`
(d) `P= n(hc)/(lambda)`
`(hc)/(lambda) =(P)/(n) =(5xx10^(-3))/(8xx10^(15))=0.625xx10^(-18) J`
`=(0.625xx10^(-18))/(1.6xx10^(-19)) =3.9 eV =E`
`K_(max) =eV_(s) =2 eV`
`E=phi + K_(max) implies 3.9 = phi + 2`
`phi =1.9 eV`
(e) `(hc)/(lambda) =phi +30`
`(hc)/(2 lambda) =phi +10`
(i)/(ii)
`2=(phi+30)/(phi+20) implies 2phi + 20=phi +30 implies phi =10 eV`
`(hc)/(lambda) =40 eV`
`lambda =(1242)/(40) =31 nm`
`lambda_(0) =(1242)/(phi(eV))nm =(1242)/(10) =124.2 nm`