Question

Question: A spectral emission line is electromagnetic radiation that is emitted

in a wavelength range narrow enough to be taken as a single wavelength. One such emission line that is important in astronomyhas a wavelength of $\mathbf{21}\mathbf{}\mathbf{cm}$ . What is the photon energy in the electromagnetic wave at that wavelength?

Short answer

The photon energy in the electromagnetic wave is$5.9\mathrm{\mu eV}$.

Step-by-step solution

Identifying the data given in the question

The wavelength of the electromagnetic wave is $\lambda =21\mathrm{cm}$

Concept used to solve the question

Photon Energy:

The energy that a single photon carries is known as photon energy. Energy is inversely correlated with wavelength because it is directly proportional to the electromagnetic frequency of the photon

Finding the photon energy

The photon energy in the electromagnetic wave can be given as

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

Where $h$is plank constant,$c$ is the velocity of light, and $\lambda$is the wavelength

We know role="math" localid="1663228885797" $hc=1240\mathrm{nm}·\mathrm{eV}$

Substituting the values

$\begin{array}{rcl}E& =& \frac{1240\mathrm{nm}·\mathrm{eV}}{21\times {10}^7\mathrm{nm}}\\ & =& 5.9\times {10}^{-6}\mathrm{eV}\\ & =& 5.9\mathrm{\mu eV}\end{array}$

Hence the photon energy in the electromagnetic wave is$5.9\mathrm{\mu eV}$.