Question

The quantum numbers of two levels involved in the emission of these photons is

• A. $4\to 2$
• B. $3\to 1$
• C. $3\to 2$
• D. $4\to 3$

Answer 1

Answer: (A)

$4\to 2$

Given, work function for Sodium is $1.82eV$ and $K.E$ of fastest electron emitted is $0.73eV$.
Total energy incident on Sodium is $2.55eV$.
Energy levels of Hydrogen are of $-13.6eV,-3.4eV,-1.51eV,0.85eV$ and so on.
Therefore, the photon corresponding to $2.55eV$ is $n=4$ to $n=2$.