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

Options

A. the work function of `A` is `2.25 eV`
B. the work function of `B` is `4.20eV`
C. `T_(A)=2.00 eV`
D. `T_(B)=2.75 eV`

Answer

A, B, C

Explanation

`K_(max)=E-W`
Therefore
`T_(A)=4.25-W_(A)`.....(i)
`T_(B)=(T_(A)-1.50)=4.70-W_(B)`.....(ii)
Equation (i) and (ii) gives,
`W_(B)-W_(A)=1.95 eV`.....(iii)
de-Broglie wavelength is given by
`lambda=(h)/(sqrt(2Km)) or lambdaprop(1)/(sqrt(K)) K=KE` of electron
`:. (lambda_(B))/(lambda_(A))=sqrt((K_(A))/(K_(B)))`
or `2=sqrt((T_(A))/(T_(A)-1.5))` or `T_(A)=2eV`