Question

Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron falls from the n = 7 to the n = 4 principal energy level. Recall that the energy levels of the H atom are given by En = –2.18 × 10–18 J(1/n2)

Answer 1

Answer:

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.

Explanation:

The energy of nth energy levels of the H atom is given as:

$E_n = -2.18 \times 10^{-18} \times \frac{1}{n^2} J$

Energy of the seventh energy level = $E_7$

$E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J$

$E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J=-4.4490\times 10^{-20} J$

Energy of the seventh energy level = $E_4$

$E_4=-2.18 \times 10^{-18} \times \frac{1}{4^2} J$

$E_4=-2.18 \times 10^{-18} \times \frac{1}{16} J=-1.3625\times 10^{-19} J$

Energy of the light emitted will be equal to the energy difference of the both levels.

$E=E_7-E_4=-4.4490\times 10^{-20} J-(-1.3625\times 10^{-19} J)$

$E=9.176\times 10^{-20} J$

Wavelength corresponding to energy E can be calculated by using Planck's equation:

$E=\frac{hc}{\lambda }$

$\lambda =\frac{hc}{E}=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{9.176\times 10^{-20} J}=2.166\times 10^{-6} m=2166 nm$

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.