Question

When photon of energy $4.0\text{eV}$ strikes the surface of a metal $A$, the ejected photoelectrons have maximum kinetic energy $T_A\text{eV}$ and de- Broglie wavelength ${\lambda }_A$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.50\text{eV}$ is $T_B=(T_A-1.5)\text{eV}$. If the de-Broglie wavelength of these photo electrons ${\lambda }_B=2{\lambda }_A$, then the work function of metal $B$ is:

• A. $4\text{eV}$
• B. $2\text{eV}$
• C. $1.5\text{eV}$
• D. $3\text{eV}$

Answer 1

The correct option is A $4\text{eV}$

de-Broglie wavelength $\left(\lambda \right)$ in terms of Kinetic energy is given by,

$\lambda =\frac{h}{\sqrt{2mK}}\Rightarrow \lambda \propto \frac{1}{\sqrt{K}}$

$\therefore \frac{{\lambda }_A}{{\lambda }_B}=\sqrt{\frac{K_B}{K_A}}$

$\Rightarrow \frac{{\lambda }_A}{{\lambda }_B}=\sqrt{\frac{T_A-1.5}{T_A}}\text{(given)}$

$\text{Also, given that}\frac{{\lambda }_A}{{\lambda }_B}=\frac{1}{2}$

On solving we get. $T_A=2\text{eV}$
$\therefore K_B=T_A-1.5=2-1.5$

$\Rightarrow K_B=0.5\text{eV}$
$\therefore$ Work function of metal $B$ is

${\varphi }_B=E_B-K_B=4.5-0.5=4\text{eV}$

Hence, option $\left(A\right)$ is correct.