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(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. |
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Answer» SOLUTION :(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` |
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