When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, `T_A` (expressed in eV) and deBroglie wavelength `lambda_A`. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20V is `T_B = T_A -1.50eV`. If the deBroglie wavelength of those photoelectrons is `lambda_B = 2lambda_A` then

Options

A. The work function of A is 1.50eV
B. The work function of B is 4.0 eV
C. `T_(A)=3.2eV`
D. All of the above

Answer

B

Explanation

From Einstein photoelectric equation, `E= phi_(0) + KE_("max")`
For metal A `4= phi_(A) + T_(A)` …(i)
For metal B `4.5 = phi_(B) + (T_(A)-1.5)` …(ii)
From Eqs (i) and (ii), we get `phi_(B)- phi_(A)=2`
Now, according to de- Broglie hypothesis, `lamda_(A)= (h)/(mv)= (h)/(sqrt(2mT_(A)))`
Similarly, `lamda_(B)= (h)/(sqrt(2mT_(B)))`
`therefore (lamda_(A))/(lamda_(B)) = sqrt((T_(B))/(T_(A))) = sqrt((T_(A)-1.5)/(T_(A)))=(1- (1.5)/(T_(A)))^(1//2)`
`((1)/(2))^(2)=1- (1.5)/(T_(A))`
On solving, `T_(A)= 2.0eV`
So, `phi_(A)= 4-T_(A)= 4-2= 2.0eV`
`phi_(B)= 6-T_(A)=6-2=4.0eV`