Question

calculate the wavelength in nanometers of the photon emitted when an electron transitions from the n=4 state to the n = 2 State in a hydrogen atom

Answer 1

Using the formula for change in energy,

$\Delta E=2.18\times10^{-18}(\frac{1}{n^2_1}-\frac{1}{n^2_2})$

where n1 and n2 are 4 and 2 respectively.

Let's calculate the change in energy when this electron makes this transition.

$\begin{gathered} \Delta E=2.18\times10^{-18}(\frac{1}{4^2}-\frac{1}{2^2}) \\ \Delta E=-4.0875\times10^{19}J \end{gathered}$

This negative value in the energy indicates that the electron jumps from a higher energy level to a lower energy level as a result of loss of energy.

From this, we can calculate the wavelength of the electron using the combination of Einstein's equation and Heisenberg's equation

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

Where h = Plank's constant

c = speed of light

substitute the values into the equation and solve for the wavelength

$\begin{gathered} E=\frac{hc}{\lambda} \\ \lambda=\frac{hc}{E} \\ \lambda=\frac{6.626\times10^{-34}\times3.0\times10^8}{4.0875\times10^{-19}} \\ \lambda=4.86\times10^{-7}m \\ \lambda=486nm \end{gathered}$

From the calculations above, the wavelength of the electron is 486nm