In an experiment on photoelectric effect photons of wavelength 300...

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 $from statement (1), 1.18=E_i-E_f\left\{E_i \& E_f \quad\right.\text{ are }\text{ energy }\text{ of }\text{ initial }\text{ and }\text{ final }\text{ orbit } \} And also from the above table, we can observe that,$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 .