When photon of energy $4.0 \mathrm{eV}$ strikes the surface of a metal $A$, the ejected photoelectrons have maximum kinetic energy $T_{A} \mathrm{eV}$ and de-Broglie wavelength $\lambda_{A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photon of energy $4.50 \mathrm{eV}$ is $T_{B}=\left(T_{A}-1.5\right) \mathrm{eV}$. If the de-Broglie wavelength of these photoelectrons $\lambda_{B}=2 \lambda_{A}$, then the work function of metal B is:
Correct Option: 1,
(1) de-Broglie wavelength $(\lambda)$,
Momentum, $m v=\frac{h}{\lambda}=p=\sqrt{2 m(K E)}$
$\therefore \lambda=\frac{h}{\sqrt{2 m K E}} \Rightarrow \lambda \propto \frac{1}{\sqrt{K E}}$
$\therefore \frac{\lambda_{A}}{\lambda_{B}}=\sqrt{\frac{K_{B}}{K_{A}}}=\sqrt{\frac{T_{A}-1.5}{T_{A}}}$ (as given)
Also, $\frac{\lambda_{A}}{\lambda_{B}}=\frac{1}{2}$
On solving we get, $T_{A}=2 \mathrm{eV}$
$\therefore K E_{B}=T_{A}-1.5=2-1.5=0.5 \mathrm{eV}$
$\therefore$ Work function of metal $B$ is
$\phi_{B}=E_{B}-K E_{B}=4.5-0.5=4 \mathrm{eV}$
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