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 funtion of B is 4.20 eV
C. `T_A=2.00eV`
D. `T_B=2.75eV`

Answer

A, B, C

Explanation

For metal `A:4.25=W_A+T_A`
Also, `T_A=(1)/(2)mv_A^2=(1)/(2)(m^2v_A^2)/(m)=(p_A^2)/(2m)(h^2)/(2mlamda_A^2)`
For metal `B:4.7=(T_A-1.5)+W_B`
Also, `T_B=(h^2)/(2mlamda_B^2)xx(2mlamda_A^2)/(h^2)=(lamda_A^2)/(lamda_B^2)`
`implies(T_A-1.5)/(T_A)=(lamda_A^2)/(2lamda_A^2)=(lamda_A^2)/(4lamda_A^2)=(1)/(4)`
`implies4T_A-6=T_AimpliesT_A=2eV`