Radiation emitted from a hydrogen discharge tube is incident on a...

The maximum energy of a photon emitted from a hydrogen discharge tube corresponds to the transition of an electron from $n = \infty$ to $n = 1$.

The energy of this incident photon is:

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

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

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

Substituting the given work function of cesium $\Phi = 1.9 \text{ eV}$:

K_{\text\max} = 13.6 - 1.9 = 11.7 \text{ eV}