CGPET 2021 Solved Paper

According to Bohr's theory, the energy required for transition of $H$-atom from $n=6$ to $n=8$ is given as $\Delta\varepsilon_6\rightarrow8 = 13.6\,\mathrm{eV} \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] = 13.6\,\mathrm{eV} \left[ \frac{1}{6^2} - \frac{1}{8^2} \right]$ $= 13.6\,\mathrm{eV} \times 0.012 = 0.1652\,\mathrm{eV}$ ........(i) (a) Energy for transition of H-atom electron from $n=5$ to $n=7$. $\Delta\varepsilon_5\rightarrow7 = 13.6\,\mathrm{eV} \left[ \frac{1}{5^2} - \frac{1}{7^2} \right]$ $= 13.6\,\mathrm{eV} \times 0.196$ $\Delta\varepsilon_5\rightarrow7 = 0.266$ Energy for transition from $n = 5$ to $n = 7$ is not equal to transition from $n=6$ to $n=8$. (b) Energy for transition from $n=7$ to $n=9$, $\Delta\varepsilon_7\rightarrow9 = 13.6\,\mathrm{eV} \left[ \frac{1}{7^2} - \frac{1}{9^2} \right]$ $= 13.6\,\mathrm{eV} \times 0.0081 = 0.1096\,\mathrm{eV}$ ∴ Energy for transition from $n=7$ to $n=9$ is not equal to transition from $n=6$ to $n=8$. (c) The transition from $n=6$ to $n=8$ has energy less than transition from $n=5$ to $n=7$. ∴ Option (c) answer is correct.