Question

Two photons of energy $2.5 \text{ eV}$ and $3.5 \text{ eV}$ fall on a metal surface of work function $1.5 \text{ eV}$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

• A. $1 : 4$
• B. $2 : 1$
• C. $1 : 2$
• D. $1 : \sqrt{2}$

Hint:

Don't confuse photon energy with kinetic energy. The work function must always be subtracted from the incident energy before calculating the velocity ratio. If $E \leq \phi$, no electrons are emitted at all.

Answer 1

The Correct Option is D

We use Einstein's photoelectric equation to find the maximum kinetic energy of the emitted electrons, which in turn gives us their velocities. Step 1: Calculate Maximum Kinetic Energy ($K_{max}$) Einstein's equation is: $K_{max} = E - \phi$, where $E$ is photon energy and $\phi$ is the work function.

• For the first photon ($E_1 = 2.5 \text{ eV}$): $$K_1 = 2.5 - 1.5 = 1.0 \text{ eV}$$

• For the second photon ($E_2 = 3.5 \text{ eV}$): $$K_2 = 3.5 - 1.5 = 2.0 \text{ eV}$$

Step 2: Relate Kinetic Energy to Velocity Kinetic energy is given by $K = \frac{1}{2}mv^2$. Therefore, velocity $v \propto \sqrt{K}$. The ratio of maximum velocities is: $$\frac{v_1}{v_2} = \sqrt{\frac{K_1}{K_2}}\lt strong\gt Step 3: Calculate the Ratio\lt /strong\gt \frac{v_1}{v_2} = \sqrt{\frac{1.0}{2.0}} = \frac{1}{\sqrt{2}}$$ $$\text{Ratio} = 1 : \sqrt{2}$$