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
Initially only radioactive element A is present which is disintegrated into another stable element B. After 6hrs ratio o...Figure shows the graph of stopping potential versus the frequency of a photosensitive metal. The plank′s constant ...Which model is used to explain nuclear fission:-...A radioactive element X disintegrates into a stable element Y with half life 2 year. A fresh sample of X is available at...An atom $\ _{3}^{8}X$ is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 pr...