When photon of energy 4.0 e V strikes the surface of a metal A , the ejected photoelectrons have maximum kinetic energy T A e V and de-Broglie wavelength λ A . The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 e V is T B = T A - 1.5 e V . If…

Question

When photon of energy 4.0 e V strikes the surface of a metal A , the ejected photoelectrons have maximum kinetic energy T A e V and de-Broglie wavelength &#955; A . The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 e V is T B = T A - 1.5 e V . If the de-Broglie wavelength of these photoelectrons &#955; B = 2 &#955; A , then the work function of metal B is:

Aicharya · Accepted Answer

Relation between de-Broglie wavelength and K.E. is
_{$λ = \frac{h}{\sqrt{2 (K E) m_{e}}} \Rightarrow λ \propto \frac{1}{\sqrt{K E}}$}

\frac{λ_{A}}{λ_{B}} = \frac{\sqrt{K E_{B}}}{\sqrt{K E_{A}}}

\Rightarrow \frac{1}{2} = \sqrt{\frac{T_{A} - 1.5}{T_{A}}} \Rightarrow T_{A} = 2 e V

∴ K E_{B} = 2 - 1.5 = 0.5 e V

ϕ_{B} = 4.5 - 0.5 = 4 e V

When photon of energy 4.0 e V strikes the surface of a metal A , the ejected photoelectrons have maximum…

Solution