Question

An electron in the n= 7 state of a hydrogen atom makes a transition to the n= 3 state and emits a photon in the process. What is the wavelength, in nm, of the photon emitted?

Answer 1

We are asked to determine the wavelength of a photon. To do that we will use the following formula:

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

Where:

$\begin{gathered} h=\text{ Plank's constant} \\ c=\text{ speed of light} \\ E_p=\text{ change in energy of the photon} \end{gathered}$

To calculate the change in energy of the photon we will use the following formula:

$E_p=\frac{2\pi^2m_eK^2Z^2e^4}{h^2}(\frac{1}{n_3^2}-\frac{1}{n_7^2})$

Where:

$\begin{gathered} m_e=\text{ mass of the electron} \\ K=electric\text{ constant} \\ Z=\text{ atomic number} \\ e=\text{ elementary charge} \\ h=\text{ plank's constant} \end{gathered}$

The mass of the electron is given by:

$m_e=9.1\times10^{-31}kg$

The electric constant is given by:

$K=9\times10^9\frac{Nm^2}{C^2}$

The atomic number of hydrogen is Z = 1.

The elementary charge has a value of:

$e=1.6\times10^{-19}C$

Planks constant is equivalent to:

$h=6.6\times10^{-34}\frac{m^2kg}{s}$

Now, we substitute the values:

$E_p=\frac{2\pi^2(9.1\times10^{-31}kg)(9\times10^9\frac{Nm^2}{C^2})(1)^2(1.6\times10^{-19}C)^4}{(6.6\times10^{-34}\frac{m^2kg}{s})^2}(\frac{1}{3^2}-\frac{1}{7^2})$

Solving the operations:

$E_p=1.99\times10^{-19}J$

Now, we substitute the value in the formula for the wavelength:

$\lambda=\frac{(6.6\times10^{-34}\frac{m^2kg}{s})(3\times10^8\frac{m}{s})}{(1.99\times10^{-19}J)}$

Solving the operation:

$\lambda=9.95\times10^{-7}m$

In nanometers the wavelength is:

$\lambda=995nm$

Therefore, the wavelength is 995 nanometers.