Question

Light of two different frequencies whose photons have energies \(1\ \text{eV}\) and \(2.5\ \text{eV}\) respectively, successively illuminate a metallic surface whose work function is \(0.5\ \text{eV}\). Ratio of maximum speeds of emitted electrons will be

• A. \(1:2\)
• B. \(1:5\)
• C. \(1:1\)
• D. \(1:4\)

Hint:

In photoelectric effect, first calculate: \[ K_{\max}=E-\phi \] Then use: \[ v\propto \sqrt{K_{\max}} \]

Answer 1

The Correct Option is A

Concept:
This question is based on Einstein's photoelectric equation. The maximum kinetic energy of emitted photoelectrons is: \[ K_{\max}=E-\phi \] where, \[ E=\text{energy of incident photon} \] \[ \phi=\text{work function of the metal} \] Also: \[ K_{\max}=\frac{1}{2}mv^2 \] So: \[ v\propto \sqrt{K_{\max}} \]

Step 1: Write the given photon energies and work function.
First photon energy: \[ E_1=1\ \text{eV} \] Second photon energy: \[ E_2=2.5\ \text{eV} \] Work function: \[ \phi=0.5\ \text{eV} \]

Step 2: Calculate maximum kinetic energy for first photon.
\[ K_1=E_1-\phi \] \[ K_1=1-0.5 \] \[ K_1=0.5\ \text{eV} \]

Step 3: Calculate maximum kinetic energy for second photon.
\[ K_2=E_2-\phi \] \[ K_2=2.5-0.5 \] \[ K_2=2.0\ \text{eV} \]

Step 4: Find the ratio of speeds.
Since: \[ K=\frac{1}{2}mv^2 \] For the same electron mass: \[ v\propto \sqrt{K} \] Therefore: \[ v_1:v_2=\sqrt{K_1}:\sqrt{K_2} \] \[ v_1:v_2=\sqrt{0.5}:\sqrt{2.0} \] Now: \[ 0.5=\frac{1}{2} \] So: \[ v_1:v_2=\sqrt{\frac{1}{2}}:\sqrt{2} \] \[ v_1:v_2=\frac{1}{\sqrt{2}}:\sqrt{2} \] Multiplying both terms by \(\sqrt{2}\): \[ v_1:v_2=1:2 \]

Step 5: Check the options.
Option (A) \(1:2\) is correct.
Option (B) \(1:5\) is incorrect because speeds are proportional to square root of kinetic energy, not directly photon energy.
Option (C) \(1:1\) is incorrect because kinetic energies are different.
Option (D) \(1:4\) is incorrect because that would be closer to kinetic energy ratio, not speed ratio. Hence, the correct answer is: \[ \boxed{(A)\ 1:2} \]