What is the wavelength of the photon emitted when an electron in a hydrogen atom which is in the initial state $n = 8$ jumps to the final state $n = 2 ? \left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } , 1 \mathrm { eV } = \right.$ $1.60 \times 10 ^ { - 19 } \mathrm {~J}$ )
A) $389 \mathrm {~nm}$
B) $610 \mathrm {~nm}$
C) $486 \mathrm {~nm}$
D) $205 \mathrm {~nm}$
E) $104 \mathrm {~nm}$

Question

Quizplus · Accepted Answer

The energy of a photon is given by $E = hf$, where $h$ is Planck's constant and $f$ is the frequency of the photon. The frequency of a photon is related to its wavelength by the equation $c = f\lambda$, where $c$ is the speed of light. Therefore, the wavelength of a photon is given by $\lambda = \frac{c}{f}$.The energy of an electron in a hydrogen atom is given by $E = -\frac{13.6}{n^2} \mathrm {~eV}$, where $n$ is the principal quantum number of the electron. Therefore, the energy of an electron in the initial state $n = 8$ is $E_i = -\frac{13.6}{8^2} \mathrm {~eV} = -0.2125 \mathrm {~eV}$. The energy of an electron in the final state $n = 2$ is $E_f = -\frac{13.6}{2^2} \mathrm {~eV} = -3.4 \mathrm {~eV}$.The energy of the photon emitted when an electron jumps from the initial state $n = 8$ to the final state $n = 2$ is $E = E_i - E_f = -0.2125 \mathrm {~eV} - (-3.4 \mathrm {~eV}) = 3.1875 \mathrm {~eV}$. The wavelength of the photon is $\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \mathrm {~J} \cdot \mathrm { s})(3.00 \times 10^8 \mathrm {~m} / \mathrm { s})}{3.1875 \mathrm {~eV}} = 389 \mathrm {~nm}$.

What Is the Wavelength of the Photon Emitted When an Electron