Question

When photons of energy $4.25 \,eV$ strike the surface of a metal $A$, the ejected photoelectrons have maximum kinetic energy $T_{A}$ (expressed in $eV$ ) and de-Broglie wavelength $\lambda$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photons of energy $4.70 \,eV$ is $T_{B}=\left(T_{A}-1.50\, eV \right)$. If the de-Broglie wavelength (in $eV$ ) of these photoelectrons is $\lambda_{B}=2 \lambda_{A}$, then find $T_{B}$ (in $\left.eV \right)$.

Answer 1

$4.25=T_{A}+\phi_{A} ; \quad 4.70=T_{B}+\phi_{B}$
$T_{B}=T_{A}-1.5$ (Given)
$\lambda=\frac{h}{\sqrt{2 m T}}$
and $\lambda_{A}=\frac{1}{2} \lambda_{B}$
$ \Rightarrow T_{A}=4 T_{B}$