The light emitted during the transition from n = 3 to n = 2 in...

The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\dfrac{13.6}{n^2} \text{ eV}$.

For the $H_\alpha$ transition from $n = 3$ to $n = 2$, the energy of the emitted photon is:

$E = E_3 - E_2$

$E = -\dfrac{13.6}{3^2} -$\left(-\dfrac{13.6}{2^2}\right)

$E = -1.51 + 3.40 = 1.89 \text{ eV}$

For photoelectric emission to take place, the energy of the incident photon must be greater than or equal to the work function of the metal ($\Phi$).

$\Phi \le E$

Therefore, the maximum work function the metal can have is \Phi_{\text\max} = 1.89 \text{ eV}.