Question

When photons of energy `4.25 eV` strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and De-broglie wavelength `lambda_(A)`. The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is `T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A) `, then
A. The work function of A is 2.25 eV
B. The work function of B is `4.20 eV`
C. `T_(A) = 2.00 eV`
D. `T_(B) = 2.75 eV`

Answer 1

Correct Answer - A::B::C
(1) `lambda_(A) = (h)/(sqrt(2mT_(A)))` , (2) `lambda_(B) = (h)/(sqrt(2mT_(B)))`
(3) `T_(B) = T_(A) - 1.5 eV` , (4) `lambda_(B) = 2lambda_(A)`