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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» (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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