Question

When photons of energy $4.0 \, eV$ fall on the surface of a metal $A$ , the ejected photoelectrons have maximum kinetic energy $T_{A}$ (in $eV$ ) and a de-Broglie wavelength $\lambda _{A}$ . When the same photons fall on the surface of another metal $B$ , the maximum kinetic energy of ejected photoelectrons is $T_{B}=T_{A}-1.5 \, eV$ . If the de-Broglie wavelength of these photoelectrons is $\lambda _{B}=2\lambda _{A}$ , then the work function of metal $B$ is

• A. $2eV$
• B. $3eV$
• C. $2.5eV$
• D. $3.5eV$

Answer 1

Answer: (D)

The relation between de-Broglie wavelength and kinetic energy is
$\lambda =\frac{h}{\sqrt{2 K m_{e}}}$
$\frac{\lambda _{A}}{\lambda _{B}}=\frac{\sqrt{K_{B}}}{\sqrt{K_{A}}}\Rightarrow \frac{1}{2}=\sqrt{\frac{T_{A} - 1 . 5}{T_{A}}}\Rightarrow T_{A}=2eV$
$\Rightarrow K_{B}=2-1.5=0.5eV$
So, the work-function of $B$ is
$\phi_{B}=4.0-0.5=3.5eV$