Question

Two spectral lines of minimum and maximum energy transitions, constituting the Balmer series, fall on two metals X and Y of work functions as given below. Which of these metals will exhibit photoelectric emission? 

a. Metal X with work function 1.7 eV. 

b. Metal Y with work function 3.1 eV.

Answer 1

Energy of photon emitted can be calculated by the formula

\(E = 13.6\left[\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right]\) 

The first Balmer spectral line (of minimum energy) emission could be due to the transition between

n₁ = 2 and n₂ = 3

The energy of this photon

\(= 13.6[\frac{1}{2^2} - \frac{1}{3^2}] = 1.9\ eV\)

As the energy of an incident photon is greater than the work function of metal X but less than the work function of metal Y, this photon can result in photoelectric emission in only metal X.

The second Balmer spectral line (of maximum energy) emission corresponds to the transition:

n₁ = 2 and n₂ = infinity

The energy of this photon

\(=13.6[\frac{1}{2^2} - \frac{1}{\infty^2}] = 3.4\ eV\) 

As the energy of the incident photon exceeds the work functions of both the metal X & Y, this photon can result in photoelectric emission in both metals X and Y.