BITSAT 2012 Paper

Given $\phi = 2.25\ \text{eV}$ $\lambda = 300 \times 100^{-3}\ \text{m}$ Energy of the given photon, $E = h v$ $= \frac{hc}{2} = \frac{663 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9}} = 6.63 \times 10^{-19}\ \text{J} = 4.13\ \text{eV}$ Now, $E = \frac{hc}{2} - \phi = 4.137 - 2.25 = 1.88\ \text{eV} \rightarrow (1)$ So, $1.88\ \text{e V}$ energy is used to jump from one orbit to another orbit by electron. Therefore, energy of different orbital of hydrogen $\Rightarrow n = 1\;\;2\;\;3\;\;4$ $$E = - 1 3 . 6\ \mathrm{e v} \& - 3 . 4 \& - 1 . 5 1 \& - 0 . 8 5$$ And from statement (1), $1 . 1 8 = E_{i} - E_{f}$ $\{ E_{i} \& E_{f}$ are energy of initial and final orbit\} And also from the above table, we can observe that, $E_{i} - E_{f} = (- 1 . 5 1) - (- 3 . 4) = 1 . 8 9\ \text{e V}$ $= E_{2} - E_{3}$ That the electron jumps from 3rd to 2nd orbit. Therefore, with one photon of $300\ \text{nm}$ wavelength hydrogen electron can jump from 3 to 2 .