# Photons of energies 4.25 eV and 4.7 eV are incident on two metal surface A and B respectively. The maximum KE of emitted electrons are respectively TA eV and TB = (TA – 1.5) eV. The ratio de-Broglie wavelengths of photo electrons from them is  λA ∶ λB = 1 : 2 , then find the work function of A and B.

Q: Photons of energies 4.25 eV and 4.7 eV are incident on two metal surface A and B respectively. The maximum KE of emitted electrons are respectively TA eV and TB = (TA – 1.5) eV. The ratio de-Broglie wavelengths of photo electrons from them is  λA ∶ λB = 1 : 2 , then find the work function of A and B.

Sol: Debroglie wavelength , $\large \lambda \propto \frac{1}{\sqrt{K}}$

$\large \frac{\lambda_B}{\lambda_A} = \sqrt{\frac{T_A}{T_B}}$

$\large 2 = \sqrt{\frac{T_A}{T_A – 1.5}}$

TA = 2 eV

WA = 4.25 – TA = 2.25 eV

TB = TA – 1.5 = 2 – 1.5 = 0.5 eV

WB = 4.7- TB = 4.7 – 0.5 = 4.2 eV.