Nuclear Physics and RadioactivityHard
Question
When photons of energy 4.25 eV strike the surface of a metal A, the ejected photo electrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. the maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA - 1.5 eV). If the de-Broglie wavelength of these electrons of these electrons are related as λB = 2λA = , then
Options
A.The work function of A is 2.25 eV
B.The work function of B is 4.20 eV
C.TA = 2.00 eV
D.TB = 2.75 eV
Solution
(1) λA = 
(2) λB = 
(3) TB = TA - 1.5 eV (4) λB = 2λA
(2) λB = (3) TB = TA - 1.5 eV (4) λB = 2λA
Create a free account to view solution
View Solution FreeMore Nuclear Physics and Radioactivity Questions
The figure shows the results of an experiment involving photoelectric effect. The graphs A, B C and D relate to a light ...Energy in an atom bomb is produced by the process of :...A photo sensitive metallic surface has work function hv0. If photons of energy 2hv0 fall on this surface, the electrons ...Binding energy per nucleon of deutron is 1.112 MeV and binding energy per nucleon of an α particle is 7.07 MeV, the...In an atom, two electrons move around the nuclei in circular orbits of radii R and 4R The. Ratio of the time taken by th...