Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition $n=3$ to $n=2$.

Answer to Problem 29P

The wavelength of the positronium for the transition $n=3$ to $n=2$ is $1313\text{nm}$.

Explanation of Solution

Formula to calculate the wavelength is,

$\lambda =\frac{1}{\mu Z^2}\left(\frac{36hc}{5k}\right)$

• $\mu$ is the reduced mass,
• $Z$ is atomic number,
• $h$ is the Planck’s constant
• $c$ is the speed of light
• $k$ is the Coulomb’s constant

Expression the wavelength for the positronium is,

${\lambda }_p=\frac{1}{{\mu }_p{Z}_p^2}\left(\frac{36hc}{5k}\right)$

• ${\mu }_p$ is the reduced mass for positronium,
• $Z_p$ is atomic number of positronium,

Expression the wavelength for the hydrogen is,

${\lambda }_H=\frac{1}{{\mu }_H{Z}_H^2}\left(\frac{36hc}{5k}\right)$

• ${\mu }_H$ is the reduced mass for hydrogen,
• $Z_H$ is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

$\begin{array}{c}\frac{{\lambda }_p}{{\lambda }_H}=\frac{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mu }_p{Z}_p^2$}\right.\right)\left(\raisebox{1ex}{$36hc$}\!\left/ \!\raisebox{-1ex}{$5k$}\right.\right)}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mu }_H{Z}_H^2$}\right.\right)\left(\raisebox{1ex}{$36hc$}\!\left/ \!\raisebox{-1ex}{$5k$}\right.\right)}\\ =\frac{{\mu }_H{Z}_H^2}{{\mu }_p{Z}_p^2}\end{array}$

Substituting, $m_e$ for ${\mu }_H$, $\left(\raisebox{1ex}{$m_e$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$ for ${\mu }_p$, $1$ for $Z_H$, $1$ for $Z_p$, $656.3\text{nm}$ for ${\lambda }_H$ to find ${\lambda }_p$,

$\begin{array}{l}\frac{{\lambda }_p}{\left(656.3\text{nm}\right)}=\frac{m_e{\left(1\right)}^2}{\left(\raisebox{1ex}{$m_e$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\left(1\right)}^2}\\ {\lambda }_p=1313\text{nm}\end{array}$

Thus, the wavelength of the positronium is $1313\text{nm}$.

Conclusion:

Therefore, the wavelength of the positronium is $1313\text{nm}$.

(b)

To determine

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$.

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition $n=3$ to $n=2$ is $164.1\text{nm}$.

Explanation of Solution

Formula to calculate the wavelength is,

$\lambda =\frac{1}{\mu Z^2}\left(\frac{36hc}{5k}\right)$

• $\mu$ is the reduced mass,
• $Z$ is atomic number,
• $h$ is the Planck’s constant
• $c$ is the speed of light
• $k$ is the Coulomb’s constant

Expression the wavelength for the helium is,

${\lambda }_{He}=\frac{1}{{\mu }_{He}{Z}_{He}^2}\left(\frac{36hc}{5k}\right)$

• ${\mu }_{He}$ is the reduced mass for Helium
• $Z_{He}$ is atomic number of Helium

Expression the wavelength for the hydrogen is,

${\lambda }_H=\frac{1}{{\mu }_H{Z}_H^2}\left(\frac{36hc}{5k}\right)$

• ${\mu }_H$ is the reduced mass for hydrogen,
• $Z_H$ is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

$\begin{array}{c}\frac{{\lambda }_{He}}{{\lambda }_H}=\frac{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mu }_{He}{Z}_{He}^2$}\right.\right)\left(\raisebox{1ex}{$36hc$}\!\left/ \!\raisebox{-1ex}{$5k$}\right.\right)}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mu }_H{Z}_H^2$}\right.\right)\left(\raisebox{1ex}{$36hc$}\!\left/ \!\raisebox{-1ex}{$5k$}\right.\right)}\\ =\frac{{\mu }_H{Z}_H^2}{{\mu }_{He}{Z}_{He}^2}\end{array}$

Substituting, $m_e$ for ${\mu }_H$, $m_e$ for ${\mu }_{He}$, $1$ for $Z_H$, $2$ for $Z_p$, $656.3\text{nm}$ for ${\lambda }_H$ to find ${\lambda }_{He}$,

$\begin{array}{l}\frac{{\lambda }_{He}}{\left(656.3\text{nm}\right)}=\frac{m_e{\left(1\right)}^2}{\left(m_e\right){\left(2\right)}^2}\\ {\lambda }_{He}=164.1\text{nm}\end{array}$

Thus, the wavelength of the Helium is $164.1\text{nm}$.

Conclusion:

Therefore, the wavelength of the Helium is $164.1\text{nm}$