Question

Photons of frequency $\nu$ are incident on the surfaces of two metals A and B of threshold frequencies $\frac{\nu}{2}$ and $\frac{\nu}{3}$. The ratio of maximum kinetic energy of electrons emitted from metal A to that from metal B is :

• A. $\frac{1}{3}$
• B. $\frac{3}{4}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Hint:

Always express maximum kinetic energy in terms of frequency fractions directly: $K_A = h\nu\left(1 - \frac{1}{2}\right) = \frac{1}{2}h\nu$ and $K_B = h\nu\left(1 - \frac{1}{3}\right) = \frac{2}{3}h\nu$. The ratio is simply $\frac{1/2}{2/3} = \frac{3}{4}$.\

Answer 1

The Correct Option is B

Concept:
• According to Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons emitted from a metal surface is given by $K_{\text{max}} = h\nu - \Phi_0$, where $\Phi_0 = h\nu_0$ is the work function of the metal.
• The threshold frequency $\nu_0$ represents the minimum frequency of incident radiation required to eject electrons from the metal surface.
• If the incident photon frequency $\nu > \nu_0$, the excess energy appears as the maximum kinetic energy of the emitted photoelectrons.

Step 1: Calculate the maximum kinetic energy for metal A
For metal surface A, the threshold frequency is given as $\nu_{0A} = \frac{\nu}{2}$.
Applying Einstein's photoelectric equation:
\[ K_A = h\nu - h\nu_{0A} \]
Substitute $\nu_{0A} = \frac{\nu}{2}$:
\[ K_A = h\nu - h\left(\frac{\nu}{2}\right) = h\nu\left(1 - \frac{1}{2}\right) = \frac{1}{2}h\nu \]

Step 2: Calculate the maximum kinetic energy for metal B
For metal surface B, the threshold frequency is given as $\nu_{0B} = \frac{\nu}{3}$.
Applying Einstein's photoelectric equation:
\[ K_B = h\nu - h\nu_{0B} \]
Substitute $\nu_{0B} = \frac{\nu}{3}$:
\[ K_B = h\nu - h\left(\frac{\nu}{3}\right) = h\nu\left(1 - \frac{1}{3}\right) = \frac{2}{3}h\nu \]

Step 3: Determine the ratio of maximum kinetic energies
Taking the ratio of $K_A$ to $K_B$:
\[ \frac{K_A}{K_B} = \frac{\frac{1}{2}h\nu}{\frac{2}{3}h\nu} \]
Cancel the common factor $h\nu$:
\[ \frac{K_A}{K_B} = \frac{1/2}{2/3} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} \]

Step 4: Conclusion
The ratio of the maximum kinetic energy of photoelectrons emitted from metal A to that from metal B is $3 : 4$, which corresponds to option (B).