Question

The graphs below show the variation of stopping potential versus frequency of incident radiation for metals A and B. 

UV radiation of appropriate wavelength is allowed to fall on both the metals. Which metal will emit photoelectrons with higher maximum kinetic energy (E_max)? Give a reason.

Answer 1

1. From the graph, the threshold frequency of Metal A is lower than that of Metal B.
2. Since work function is given by Φ=hf0\Phi = hf_0Φ = h f0​, Metal A has a smaller work function than Metal B.
3. According to Einstein’s photoelectric equation, K_max = hf − Φ.
4. The incident radiation is the same for both metals, so the value of hfhfhf remains constant.
5. As Metal A has a smaller work function, the value of K_max​ will be greater for Metal A.

Therefore, Metal A emits photoelectrons with higher maximum ki netic energy.

•  

Based on the graph and Einstein’s photoelectric equation, here is the solution Metal A will emit photoelectrons with higher maximum kinetic energy (K_max).

 

1. Threshold Frequency (f₀): The graph shows that the threshold frequency for Metal A (f_0A) is lower than that for Metal B (f_0B).
2. Work Function (Φ): The work function is directly proportional to the threshold frequency (Φ = h f₀). Since f_0A < f_0B, Metal A has a smaller work function than Metal B.
3. Einstein’s Equation: According to the formula K_max = hf − Φ, the maximum kinetic energy is the energy of the incident light minus the work function.
4. Constant Energy (hf): Since the incident UV radiation is the same for both metals, the metal with the smaller work function (Metal A) leaves more energy available for kinetic motion.

Therefore, Metal A results in a higher K_max for its emitted photoelectrons.