Given $\phi=2.25 \mathrm{eV}$
$$
\lambda=300 \times 100^{-3} \mathrm{~m}
$$
Energy of the given photon,
$$
\mathrm{E}=\mathrm{hv}
$$
$$
=\frac{\mathrm{hc}}{\lambda}=\frac{663 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}}=6.63 \times 10^{-19} \mathrm{~J}=4.13 \mathrm{eV}
$$
Now,
$$
\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi=4.137-2.25=1.88 \mathrm{eV} \rightarrow(1)
$$
So, $1.88 \mathrm{eV}$ energy is used to jump from one orbit to another orbit by electron. Therefore, energy of different orbital of hydrogen
$$
\begin{array}{rrrr}
\Rightarrow \mathrm{n}=1 & 2 & 3 & 4 \\
\mathrm{E}=-13.6 \mathrm{ev} & -3.4 & -1.51 & -0.85
\end{array}
$$
And from statement (1),
$$
1.18=E_i-E_f
$$
$\left\{E_i \& E_f \quad\right.$ are energy of initial and final orbit $\}$ And also from the above table, we can observe that,
$$
E_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}}=(-1.51)-(-3.4)=1.89 \mathrm{eV}
$$
$$
=\mathrm{E}_2-\mathrm{E}_3
$$
That the electron jumps from 3rd to 2nd orbit. Therefore, with one photon of 3oonm wavelength hydrogen electron can jump from 3 to 2 .