A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength $\frac{\lambda }{2},$ then the maximum kinetic energy of ejected electrons will be 

(The work function of metal is 1 eV) 

Given Data:

• Wavelength of first light source: $\lambda_1 = \lambda$
• Maximum kinetic energy of ejected electrons: $K_{1,\text{max}} = 2$ eV
• Work function of metal: $\phi = 1$ eV
• Wavelength of second light source: ${\lambda }_2=\frac{\lambda }{2}$

Step 1: Using Einstein’s Photoelectric Equation

The photoelectric equation is:

$$K_{\text{max}} = h \nu - \phi$$

where:

• $h \nu$ is the energy of the incident photons, given by $h \nu = \frac{hc}{\lambda}$,
• $\phi$ is the work function of the metal.

Thus,

$$K_{\text{max}} = \frac{hc}{\lambda} - \phi$$

For the first light source:

$$2 = \frac{hc}{\lambda} - 1$$

 $$\frac{hc}{\lambda} = 3 \text{ eV} \quad \text{(Equation 1)}$$

For the second light source:

$$K_{2,\text{max}} = \frac{hc}{\lambda_2} - 1$$

Using Equation 1, we express $hc$ in terms of $\lambda$:

$$\frac{hc}{\lambda} = 3 \Rightarrow hc = 3\lambda$$

Substituting into the equation for $K_{2,\text{max}}$:

${\mathrm{K}}_{2,\text{max}}=\frac{3\mathrm{\lambda }}{{\displaystyle \raisebox{1ex}{$\mathrm{\lambda }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-1=5 \mathrm{eV}$