Chemistry - Atomic Structure Question with Solution | TestHub

When photons of energy 4.25 eV strikes the surface of a metal 'A', the ejected photoelectrons have maximum kinetic energy ${\mathrm{T}}_{\mathrm{A}}$ (in eV ) and de Broglie wavelength ${\lambda }_{\mathrm{A}}$. The maximum kinetic energy of photoelectrons liberated from another metal ' B ' by photons of energy 4.20 eV is ${\mathrm{T}}_{\mathrm{B}}\left(={\mathrm{T}}_{\mathrm{A}}-1.50\mathrm{eV}\right)$. If the de Broglie wave length of these photoelectrons is ${\lambda }_B\left(=2{\lambda }_A\right)$, then