When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, `T_(A)` (expressed in eV) and de Broglie wavelength `lambda_(A)`. The mximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is `T_(B) = T_(A) - 1.50 eV`. If the de Broglie wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A)`, then which is not correct :

Options

A. The work function of A is 2.25 eV
B. The work function of B is 3.70 eV
C. `T_(A) = 2.00 eV`
D. `T_(B) = 2.75 eV`

Answer

D