If the wavelengths of H_α and H_ß, the first two lines of the hydrogen Balmer series, are 656.3 nm and 486.2 nm, respectively, what is the wavelength of H_γ, the third line of the Balmer series? A)486.2 nm B)1875 nm C)364.6 nm D)434.1 nm

Question

Quizplus · Accepted Answer

The Balmer series formula is given by: $\frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{n^2}ight)$, where $R_H$ is the Rydberg constant for hydrogen and $n$ is the energy level of the electron. We are given the wavelengths of the first two lines, which correspond to $n=3$ and $n=4$:

$\frac{1}{\lambda_1}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}ight)$

$\frac{1}{\lambda_2}=R_H\left(\frac{1}{2^2}-\frac{1}{4^2}ight)$

We can solve these equations for $R_H$ by taking the reciprocal of both sides and subtracting them:

$\frac{1}{\lambda_2}-\frac{1}{\lambda_1}=R_H\left(\frac{1}{4^2}-\frac{1}{3^2}ight)$

$R_H=\frac{(\frac{1}{\lambda_2}-\frac{1}{\lambda_1})}{(\frac{1}{4^2}-\frac{1}{3^2})}=\frac{1}{364.6	ext{ nm}}$

Now we can use this value of $R_H$ to find the wavelength of the third line of the Balmer series, which corresponds to $n=5$:

$\frac{1}{\lambda_3}=R_H\left(\frac{1}{2^2}-\frac{1}{5^2}ight)$

$\lambda_3=\frac{1}{R_H\left(\frac{1}{2^2}-\frac{1}{5^2}ight)}=434.1	ext{ nm}$

If the Wavelengths of Hα and Hß, the First Two

If the Wavelengths of H_α and H_ß, the First Two