91Ó°ÊÓ

We are given Equation 6.5 for the energy levels in hydrogen atom, which is: \( E_n = -\frac{13.6\,\text{eV}}{n^2} \) where \(E_n\) is the energy of an electron at a particular energy level n.

For n = 2, \(E_2 = -\frac{13.6\,\text{eV}}{(2)^2} = -\frac{13.6\,\text{eV}}{4} = -3.4\,\text{eV}\) For n = 6, \(E_6 = -\frac{13.6\,\text{eV}}{(6)^2} = -\frac{13.6\,\text{eV}}{36} \approx -0.378\,\text{eV}\)

The energy difference ∆E between these two levels is given by: ∆E = E_6 - E_2 ≈ -0.378 eV - (-3.4 eV) ≈ 3.022 eV

The energy difference (∆E) is related to the wavelength of the released radiation (λ) by the formula: \( \Delta E = \frac{hc}{\lambda} \) where h is the Planck's constant (\(6.63\times10^{-34}\,\text{Js}\)), and c is the speed of light (\(3\times10^8\,\text{m/s}\)). Rearrange the formula to solve for λ: \( \lambda = \frac{hc}{\Delta E} \) To calculate the wavelength, we need to convert ∆E to joules: \( \Delta E = 3.022\,\text{eV} \times \frac{1.6\times10^{-19}\,\text{J}}{1\,\text{eV}} \approx 4.83\times10^{-19}\,\text{J} \) Now, we can calculate the wavelength λ: \( \lambda = \frac{(6.63\times10^{-34}\,\text{Js})(3\times10^8\,\text{m/s})}{4.83\times10^{-19}\,\text{J}} \approx 4.11\times10^{-7}\,\text{m} \)

The visible region of the electromagnetic spectrum ranges approximately from \(4\times10^{-7}\,\text{m}\) (violet) to \(7\times10^{-7}\,\text{m}\) (red). Since the calculated wavelength \(4.11\times10^{-7}\,\text{m}\) lies within this range, it is in the visible region. This wavelength is close to the blue-violet boundary, so the color of the released radiation would appear as bluish-violet. In conclusion, the energies of electrons in the hydrogen atom are -3.4 eV for n=2 and -0.378 eV for n=6. The wavelength of the released radiation when an electron moves from n=6 to n=2 is \(4.11\times10^{-7}\,\text{m}\), and its color is bluish-violet.