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 4.20 eV
C. `T_(A) = 2.00 eV`
D. `T_(B) = 2.75 eV`

Answer

A, B, C

Explanation

`K_(max)=E-V`
Therefore
`T_(A) = 4.25 - W_(A)`….(i)
`T_(B) = (T_(A) - 1.50) = 4.70 - W_(B)` …(ii)
Equation (i) and (ii)
`W_(B) - W_(A) = 1.95 eV` …(iii)
de-Broglie wavelength is given by
`lamda = (h)/(sqrt(2Km))` or `lamda = (1)/(sqrt(K))K=KE` of electron
`:. (lamda_(B))/(lamda_(A))=sqrt((K_(A))/(K_(B)))`
or `2 = sqrt((T_(A))/(T_(A)-1.5))` or `T_(A) = 2eV`
From equation (i)
`W_(B) + 4.25 - T_(A) = 2.25 eV`
From equations (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`.