When photon of energy `25eV` strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy `T_(A)eV` and de Brogle wavelength `lambda_(A)` .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy `4.76 eV` is `T_(B) = (T_(A) = 1.50) eV` .If the de broglie wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A)` then
i.` (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV`
III`T_(A) = 2.0 eV IV. T_(B) = 3.5 eV`

Options

A. I,II
B. II,III,IV
C. I,II,III
D. I,II,III,IV

Explanation

`W_(A) + T_(A) = 4.25 rArr W_(A) = 2.25 eV`
`W_(A) + T_(B) = 4.7 rArr W_(B) = 4.2 eV`
`lambda_(B) = 2 lambda_(A)`
`rArr KE_(A) = 4KE_(B)`
`rArr T_(A) = 4(T_(B))`
`rArr T_(A) = 2.0 eV`
`T_(A) = 0.5 eV`
Hence option c