When photons of energy 4.25 eV strike the surface of metal $A$, the ejected photoelectrons have maxi

Question

When photons of energy 4.25 eV strike the surface of 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 photons of energy 4.70 eV is $T_{B}=\left(T_{A}-1.50ight) \mathrm{eV}$. If the de Broglie wavelength of these photoelectrons is $\lambda_{B}=2 \lambda_{A}$, then\

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(a, b, c) : Consider metal $A$ Incident energy $=$ work function + Kinetic energy of photoelectrons $$\begin{equation*} 	herefore \quad 4.25(\mathrm{eV})=W_{A}+T_{A} 	ag{i} \end{equation*}$$ Kinetic energy $=\frac{p^{2}}{2 m}$\ where $p=$ momentum\ $	herefore \quad T_{A}=\frac{p_{A}^{2}}{2 m}=\frac{1}{2 m}\left(\frac{h}{\lambda_{A}}ight)^{2}$,\ by de Broglie equation $$\begin{equation*} 	herefore \quad T_{A}=\frac{1}{2 m}\left(\frac{h}{\lambda_{A}}ight)^{2} 	ag{ii} \end{equation*}$$ Consider metal $B$ : $$\begin{equation*}7=\left(T_{A}-1.5ight)+W_{B} 	ag{iii} \end{equation*}$$ Also $T_{B}=\frac{1}{2 m}\left(\frac{h}{\lambda_{B}}ight)^{2}$\ From (iv) and (ii), we get\ $\frac{T_{B}}{T_{A}}=\left(\frac{\lambda_{A}}{\lambda_{B}}ight)^{2}$ or $\frac{T_{A}-1.5}{T_{A}}=\left(\frac{\lambda_{A}}{\lambda_{B}}ight)^{2}$\ $\left[\because T_{B}=T_{A}-1.5ight]$\ or $\frac{T_{A}-1.5}{T_{A}}=\left(\frac{\lambda_{A}}{2 \lambda_{A}}ight)^{2} \quad\left[\because \lambda_{B}=2 \lambda_{A}ight]$\ or $\quad 4 T_{A}-6.0=T_{A}$ or $3 T_{A}=6$\ or $\quad T_{A}=2.00 \mathrm{eV}$\ From (i), $$\begin{equation*} =4.25-2=2.25 \mathrm{eV} 	ag{vi} \end{equation*}$$ From (iii), $W_{B}=4.7-\left(T_{A}-1.5ight)$ $$ =4.7-2+1.5 $$ $$\begin{equation*} =4.20 \mathrm{eV} 	ag{vii} \end{equation*}$$ Again $T_{B}=T_{A}-1.5=2-1.5=0.5 \mathrm{eV}$\ Options (a), (b), (c) are correct as depicted in equations (vi), (vii) and (v) above.\