A hydrogen atom makes a downward transition from the $n = 15$ state to the $n = 5$ state. Find the wavelength of the emitted photon. $\left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right.$ )
A) $2.56 \mu \mathrm { m }$
B) $2.05 \mu \mathrm { m }$
C) $3.07 \mu \mathrm { m }$
D) $1.54 \mu \mathrm { m }$

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 $f = c/\lambda$, where $c$ is the speed of light and $\lambda$ is the wavelength of the photon. The energy of a photon is also related to its wavelength by $E = hc/\lambda$.The energy of a hydrogen atom in the $n$th state is given by $E = -13.6 \mathrm {~eV} / n^2$. The energy of a photon emitted when a hydrogen atom makes a transition from the $n_i$th state to the $n_f$th state is given by $\Delta E = E_i - E_f = -13.6 \mathrm {~eV} \left(1/n_f^2 - 1/n_i^2\right)$.In this problem, the hydrogen atom makes a transition from the $n = 15$ state to the $n = 5$ state. The energy of the photon emitted is $\Delta E = -13.6 \mathrm {~eV} \left(1/5^2 - 1/15^2\right) = 1.89 \mathrm {~eV}$. The wavelength of the photon is $\lambda = hc/\Delta E = (6.626 \times 10^{-34} \mathrm {~J} \cdot \mathrm { s})(3.00 \times 10^8 \mathrm {~m} / \mathrm { s})/(1.89 \mathrm {~eV})(1.60 \times 10^{-19} \mathrm {~J} / \mathrm {~eV}) = 2.56 \mu \mathrm { m }$.Therefore, the correct answer is A.

A Hydrogen Atom Makes a Downward Transition from The n=15n = 15n=15

A Hydrogen Atom Makes a Downward Transition from The $n = 15$