If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission?

Choose the correct answer from the options given below:

Na and K only

The correct answer is Option (1) → Na and K only

Energy of incident photon:

$ E = \frac{hc}{\lambda} = \frac{1240 \ \text{eV·nm}}{412.5 \ \text{nm}} $

$ E = 3.01 \ \text{eV} $

Compare with work functions:

Na : $1.92 \ \text{eV} \lt 3.01 \ \text{eV} $ → emission occurs.

K : $2.15 \ \text{eV} \lt 3.01 \ \text{eV} $ → emission occurs.

Ca : $3.20 \ \text{eV} \gt 3.01 \ \text{eV} $ → no emission.

Mo : $4.17 \ \text{eV} \gt 3.01 \ \text{eV} $ → no emission.

Metals showing photoelectric emission: Na and K.