Question

The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H. [Note: These are the electronic hydrogen and deuterium atoms, not the muonic forms.]

Answer 1

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

$\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})$

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

$R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}$

Here, $R_{\infty}$ is the "general" Rydberg constant, $m_e$ is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, $M=1.67*10^{-27}kg$:

$R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}$

Now, we calculate the wavelength for hydrogen:

$\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm$

For deuterium, we have $M=2(1.67*10^{-27}kg)$:

$R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm$