When two monochromatic lights of frequency, $\mathrm{\nu }$ and $\frac{\mathrm{\nu }}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{{\mathrm{V}}_{\mathrm{s}}}{2}\text{ and }{\mathrm{V}}_{\mathrm{s}}$ respectively. The threshold frequency for this metal is:

Complete Solution:

Given:

Two frequencies: ν and ν₂ = 2ν

Stopping potentials: V_s and V_s2

We need to determine the threshold frequency for the metal.

Einstein’s Photoelectric Equation:

eV_s = h(ν - ν₀)

Where:

e is the electron charge

h is Planck's constant

ν₀ is the threshold frequency

For Frequencies ν and ν₂:

For frequency ν and stopping potential V_s: eV_s = h(ν - ν₀)

For frequency ν₂ = 2ν and stopping potential V_s2: eV_s2 = h(2ν - ν₀)

Solving for Threshold Frequency:

Subtracting the two equations: e(V_s2 - V_s) = hν From this, ν = e(V_s2 - V_s) / h

Substituting back to find ν₀, we get: ν₀ = 3/2 ν

Final Answer:

The threshold frequency for the metal is 3/2 ν.