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 - B::C::D
`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 `lambda prop h/sqrtK`
`K=KE` of electron
`therefore lambda_(B)/lambda_(A)=sqrt(K_(A)/K_(B))` or `2=sqrt((T_(A))/(T_(A)-1.5))` or `T_(A)=2eV`
From equation (i) `W_(A)=4.25-T_(A)=2.25 eV`
From equation (iii), `W_(B)+1.95 eV=(2.25+1.95) eV` or `W_(B)=4.20 eV`.
`T_(B)=4.70-W_(B)=4.70-4.20=0.50 eV`.