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=(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=(2\pi^2m_eK^2Z^2e^4)/(h^2)((1)/(n_3^2)-(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*10^(-31)kg`

The electric constant is given by:

`K=9*10^9(Nm^2)/(C^2)`

The atomic number of hydrogen is Z = 1.

The elementary charge has a value of:

`e=1.6*10^(-19)C`

Planks constant is equivalent to:

`h=6.6*10^(-34)(m^2kg)/(s)`

Now, we substitute the values:

`E_p=(2\pi^2(9.1*10^(-31)kg)(9*10^9(Nm^2)/(C^2))(1)^2(1.6*10^(-19)C)^4)/((6.6*10^(-34)(m^2kg)/(s))^2)((1)/(3^2)-(1)/(7^2))`

Solving the operations:

`E_p=1.99*10^(-19)J`

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

`\lambda=((6.6*10^(-34)(m^2kg)/(s))(3*10^8(m)/(s)))/((1.99*10^(-19)J))`

Solving the operation:

`\lambda=9.95*10^(-7)m`

In nanometers the wavelength is:

`\lambda=995nm`

Therefore, the wavelength is 995 nanometers.