Radiation from a hydrogen-discharge tube passes through a filter...

The energy of photons emitted by a hydrogen atom is given by:

$E = 13.6 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) \text{ eV}$

The corresponding wavelength is $\lambda = \dfrac{1240 \text{ eV nm}}{E}$.

Let us check the transitions for the Balmer series ($n_1 = 2$):

For $n_2 = 3 \to n_1 = 2$:
$E = 13.6 \left( \dfrac{1}{4} - \dfrac{1}{9} \right) = 1.89 \text{ eV}$
$\lambda = \dfrac{1240}{1.89} \approx 656 \text{ nm}$

For $n_2 = 4 \to n_1 = 2$:
$E = 13.6 \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = 2.55 \text{ eV}$
$\lambda = \dfrac{1240}{2.55} \approx 486 \text{ nm}$

For $n_2 = 5 \to n_1 = 2$:
$E = 13.6 \left( \dfrac{1}{4} - \dfrac{1}{25} \right) = 2.86 \text{ eV}$
$\lambda = \dfrac{1240}{2.86} \approx 434 \text{ nm}$

The filter transmits only wavelengths $\lambda > 440 \text{ nm}$. Thus, the $n=5 \to 2$ transition and all higher energy transitions (including the entire Lyman series, which has $\lambda < 122 \text{ nm}$) are blocked.

The transmitted photons are from the $n=3 \to 2$ and $n=4 \to 2$ transitions (as well as higher series like Paschen, which have energies $< 1.51 \text{ eV}$).

The work function of the metal is $\Phi = 2.0 \text{ eV}$.

Since the energy of the $n=3 \to 2$ photon ($1.89 \text{ eV}$) and all higher series photons is less than the work function, they cannot cause photoelectric emission.

The only transmitted photon capable of causing photoelectric emission is from the $n=4 \to 2$ transition, with $E = 2.55 \text{ eV}$.

Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is:

K_{\text\max} = E - \Phi

K_{\text\max} = 2.55 \text{ eV} - 2.0 \text{ eV} = 0.55 \text{ eV}

The stopping potential $V_s$ is given by eV_s = K_{\text\max}, which yields:

$V_s = 0.55 \text{ V}$