Question

When photoelectrons of energy $4.25eV$ strike the surface of metal $A$, the ejected photoelectrons have maximum kinetic energy $T_{A}eV$ and de-Broglie wavelength ${\lambda }_A$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photons of energy $4.70eV$ is $T_B=(T_A-1.5)eV$. If de-Broglie wavelength of these photoelectrons is ${\lambda }_B=2{\lambda }_A$, then

• A. Work function of $A$ is $2.25eV$.
• B. Work function of $B$ is $4.20eV$.
• C. $T_A=2.00eV$
• D. $T_B=2.75eV$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Work function of $A$ is $2.25eV$.
B Work function of $B$ is $4.20eV$.
C $T_A=2.00eV$

Using de-Broglie wave equation,
$p_A=\frac{h}{{\lambda }_A}$, $T_A=\frac{p_A^2}{2m}=\frac{h^2}{2m{\lambda }_A^2}$, $T_B=\frac{h^2}{2m{\lambda }_B^2}=\frac{h^2}{8m{\lambda }_A^2}$

Using work function equation,
$\frac{h^2}{2m{\lambda }_A^2}=4.25-{\varphi }_A\dots \dots \left(1\right)$
$\frac{h^2}{8m{\lambda }_A^2}=4.70-{\varphi }_B\dots \dots \left(2\right)$
Also, $\frac{h^2}{8m{\lambda }_A^2}=\frac{h^2}{2m{\lambda }_A^2}-1.5$
$\frac{3}{8}\frac{h^2}{m{\lambda }_A^2}=1.5\Rightarrow \frac{h^2}{m{\lambda }_A^2}=4eV$
$\therefore T_A=2eV,T_B=0.5eV$
${\varphi }_A=2.25eV,{\varphi }_B=4.70-0.5=4.20eV$