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

#### Solution:

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

(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:

(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{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:

(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

#### Final Answer: