Photoelectrons are ejected from a cesium surface ( = 1.9 eV ) by a...

The minimum kinetic energy required by a photoelectron to ionize a hydrogen atom in its ground state is equal to the ionization energy of hydrogen, which is $13.6 \text{ eV}$.

According to Einstein's photoelectric equation, the maximum kinetic energy of the ejected photoelectrons is given by

K_{\text\max} = \dfrac{hc}{\lambda} - \Phi

For the photoelectrons to be able to ionize the hydrogen atoms, their kinetic energy must be at least $13.6 \text{ eV}$. Therefore,

$\dfrac{hc}{\lambda} - \Phi \ge 13.6 \text{ eV}$

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

$\dfrac{hc}{\lambda} - 1.9 \text{ eV} \ge 13.6 \text{ eV}$

$\dfrac{hc}{\lambda} \ge 15.5 \text{ eV}$

The maximum wavelength corresponds to the minimum energy required:

\lambda_{\text\max} = \dfrac{hc}{15.5 \text{ eV}}

Using $hc = 1240 \text{ eV nm}$,

\lambda_{\text\max} = \dfrac{1240}{15.5} \text{ nm} = 80 \text{ nm}