# Calculate the longest wavelength of the electromagnetic radiation

Calculate the longest wavelength of the electromagnetic radiation emitted by the hydrogen atom in undergoing a transition from the n = 7 level.

#### Solution:

The electron will go from n = 7 to the n = 6 level. We will use the following equation to figure out the frequency.

$E=hv$

(Where $E$ is energy, $h$ is Planck’s constant, and $v$ is the frequency)

Since we know the electron undergoes a transition from n = 7 to n = 6, we can write the following:

$E_7\,-\,E_6\,=\,\dfrac{-R_H}{7^2}\,-\,\dfrac{-R_H}{6^2}$

(simplify by finding a common denominator)

$\dfrac{-R_H}{49}\,-\,\dfrac{-R_H}{36}\,=\,\dfrac{13R_H}{1764}$

We can now write the frequency emitted:

$v\,=\,\dfrac{E}{h}$

$v\,=\,\dfrac{13R_H}{1764h}$

$v\,=\,\dfrac{13}{1764}\times \dfrac{2.179\times 10^{-18}\text{J}}{6.626\times 10^{-34}\text{J}\cdot \text{s}}$

$v\,=\,2.42\times 10^{13}$ /s

We can now figure out the wavelength by using the following equation:

$\lambda\,=\,\dfrac{c}{v}$

(Where $\lambda$ is the wavelength, $c$ is the speed of light, $v$ is the frequency)

$\lambda\,=\,\dfrac{2.998\times 10^8\text{m/s}}{2.42\times 10^{13}/s}$

$\lambda\,=\,1.238\times 10^{-5}$ m

The longest wavelength is $1.238\times 10^{-5}$ m