Question

Photoemission from a metal surface takes place for frequencies $\nu_1$ and $\nu_2$ of incident rays ($\nu_1 > \nu_2$). The maximum kinetic energy of the photoelectrons emitted is in the ratio $1 : K$. The threshold frequency ($\nu_0$) of the metallic surface is

• A. $\frac{K\nu_2 - \nu_1}{K - 1}$
• B. $\frac{\nu_1 - \nu_2}{K - 1}$
• C. $\frac{\nu_2 - \nu_1}{K}$
• D. $\frac{K\nu_1 - \nu_2}{K - 1}$

Hint:

To verify your algebraic steps quickly during an exam, check the dimensions or units of your final expression. The denominator $(K - 1)$ is a dimensionless scalar ratio, while the numerator $(K\nu_1 - \nu_2)$ carries units of frequency ($\text{s}^{-1}$). This confirms the final expression has the correct physical units for frequency!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question concerns the photoelectric effect across two independent monochromatic illumination setups on the same target metal. We are given the two incident light frequencies ($\nu_1$ and $\nu_2$) and the ratio of the corresponding maximum kinetic energies ($1 : K$). We need to derive an expression for the threshold frequency ($\nu_0$).

Step 2: Key Formula or Approach:
According to

Einstein's Photoelectric Equation, the maximum kinetic energy ($K.E.$) of an emitted photoelectron is given by: $$K.E. = h\nu - h\nu_0 = h(\nu - \nu_0)$$ Where $h$ is Planck's constant, $\nu$ is the incident frequency, and $\nu_0$ is the threshold frequency of the metal surface.

Step 3: Detailed Explanation:
Set up individual energy equations for both frequency cases: 1. For incident light of frequency $\nu_1$: $$(K.E.)_1 = h(\nu_1 - \nu_0)$$ 2. For incident light of frequency $\nu_2$: $$(K.E.)_2 = h(\nu_2 - \nu_0)$$ We are given the ratio of these kinetic energies as $\frac{(K.E.)_1}{(K.E.)_2} = \frac{1}{K}$. Divide the first equation by the second equation: $$\frac{h(\nu_1 - \nu_0)}{h(\nu_2 - \nu_0)} = \frac{1}{K}$$ The constant $h$ cancels out from the numerator and denominator: $$\frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = \frac{1}{K}$$ Cross-multiply to clear the fractions and solve for $\nu_0$: $$K(\nu_1 - \nu_0) = 1(\nu_2 - \nu_0)$$ $$K\nu_1 - K\nu_0 = \nu_2 - \nu_0$$ Group all terms containing the threshold frequency $\nu_0$ on one side of the equation: $$K\nu_1 - \nu_2 = K\nu_0 - \nu_0$$ Factor out $\nu_0$ on the right side: $$K\nu_1 - \nu_2 = \nu_0(K - 1)$$ Isolate $\nu_0$ by dividing both sides by $(K - 1)$: $$\nu_0 = \frac{K\nu_1 - \nu_2}{K - 1}$$

Step 4: Final Answer:
The threshold frequency of the metallic surface is $\frac{K\nu_1 - \nu_2}{K - 1}$, which matches option (D).