A light of wavelength ‘λ’ is incident on a metal having work function φ = 3.4 eV. The stopping potential measured for th…

The energy of the incident photon (\(E\)) is related to the work function and the stopping potential by:

\(E=ϕ+eV\)E

Here:

• \(ϕ=3.4\text{ }\text{eV}\)
• \(eV=1.6\text{ }\text{eV}\)
• The total energy of the incident photon:

\(E=3.4+1.6=5.0\text{ }\text{eV}\)

The energy of the photon is also given by:

\(E=\frac{hc}{\lambda }\)​

Rearranging for \(\lambda\):

\(\lambda =\frac{hc}{E}\)​

Substitute the given values:

\(\lambda =\frac{12400}{5}=2480\text{ }\overset{˚}{\text{A}}\)

Convert to nanometers (since \(1\text{ }\overset{˚}{\text{A}}=0.1\text{ }\text{nm}\)1A˚=0.1nm):

\(\lambda =248\text{ }\text{nm}\)λ=248nm

Correct option (b) 248nm