+2 votes
in Physics by kratos

When photons of energy 4.25 eV strike the surface of a metal A the ejected photo electrons have a maximum kinetic energy EA eV and de Broglie wavelength λA. The maximum kinetic energy of photo electrons liberated from another metal B by photons of energy 4.70 eV is EB = (EA – 1.50) eV. If the de Broglie wavelength of these photo electrons is λB = 2λA, then

(a) the work function of A is 2.25 eV

(b) the work function of B is 4.20 eV

(c) EA = 2.0 eV

(d) All of these

1 Answer

+6 votes
by kratos
 
Best answer

Correct option (d)

Explanation:

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