Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=3 \rightarrow n=2\) b. \(n=4 \rightarrow n=2\) c. \(n=2 \rightarrow n=1\)

The energy level of a hydrogen atom for an electron in a state with principal quantum number n is given by the formula: \(E_n = -\frac{13.6 eV}{n^2}\) To calculate the energy released during a transition, first, we need to calculate the difference in energy levels (∆E) between the two energy levels, then convert the energy difference into the emitted wavelength using the formula: \(E = h \cdot c / \lambda\) Where E is the energy, h is the Planck's constant (4.135 x 10^-15 eV·s), c is the speed of light (3 x 10^8 m/s), and λ (lambda) is the wavelength of the light emitted. For each transition, we will calculate the energy difference and then find the corresponding wavelength of the emitted light using the constants Planck's constant and the speed of light.

Using the energy level formula, calculate the energy levels for n=3 and n=2: \(E_3 = -\frac{13.6 eV}{(3)^2} = -\frac{13.6 eV}{9} = -1.51 eV\) \(E_2 = -\frac{13.6 eV}{(2)^2} = -\frac{13.6 eV}{4} = -3.4 eV\) Now, we can calculate the energy difference (∆E) for the transition: \(\Delta E = E_3 - E_2 = -1.51 eV - (-3.4 eV) = 1.89 eV\)

Now we will use the energy difference and the constants to find the emitted wavelength: \(1.89 eV = \frac{4.135 \times 10^{-15} eV\cdot s \cdot 3\times10^8 m/s}{\lambda}\) Solving for λ, \(\lambda = \frac{4.135 \times 10^{-15} eV\cdot s \cdot 3\times10^8 m/s}{1.89 eV} = 6.56 \times 10^{-7} m\) Converting the wavelength to nm (1 m = 10^9 nm): \(\lambda = 656 nm\)

With the calculated wavelength of 656 nm, which falls in the visible light range (400-700nm), we can identify that the electromagnetic radiation emitted for transition a is a visible light.

Repeat the above calculations for the remaining transitions and identify the type of electromagnetic radiation for each. Transition b: 1. Energy difference: ∆E = 2.55 eV 2. Wavelength: λ = 486 nm 3. Type of electromagnetic radiation: Visible light Transition c: 1. Energy difference: ∆E = 10.2 eV 2. Wavelength: λ = 122 nm 3. Type of electromagnetic radiation: Ultraviolet light In conclusion, for transition a (\(n=3 \rightarrow n=2\)), the emitted light has a wavelength of 656 nm and falls in the visible light range; for transition b (\(n=4 \rightarrow n=2\)), with a wavelength of 486 nm, it is also visible light; and for transition c (\(n=2 \rightarrow n=1\)), the emitted light has a wavelength of 122 nm, which falls in the ultraviolet light range.