Question

reduced mass of hydrogen atom and deuterium

Answer 1

The reduced mass of a deuterium atom is greater than the reduced mass of a hydrogen atom.

Answer 2

The reduced mass ($\mu$) for a two-body system with masses $m_1$ and $m_2$ is given by the formula:

$$\mu = \frac{m_1 m_2}{m_1 + m_2}$$

1. Reduced mass of a Hydrogen atom ($\mu_H$):

A hydrogen atom consists of an electron ($m_e$) and a proton ($m_p$).

$$\mu_H = \frac{m_e m_p}{m_e + m_p}$$

2. Reduced mass of a Deuterium atom ($\mu_D$):

A deuterium atom consists of an electron ($m_e$) and a deuteron ($m_d$). A deuteron is the nucleus of deuterium, composed of one proton and one neutron.

$$\mu_D = \frac{m_e m_d}{m_e + m_d}$$

3. Comparison of $\mu_H$ and $\mu_D$:

We know that the mass of a deuteron ($m_d$) is approximately twice the mass of a proton ($m_d \approx 2m_p$), which means $m_d > m_p$.

To compare $\mu_H$ and $\mu_D$, we can rewrite the reduced mass formula as:

$$\mu = \frac{m_e M}{m_e + M} = \frac{m_e}{1 + \frac{m_e}{M}}$$

where $M$ is the mass of the nucleus (either $m_p$ for hydrogen or $m_d$ for deuterium).

For hydrogen:

$$\mu_H = \frac{m_e}{1 + \frac{m_e}{m_p}}$$

For deuterium:

$$\mu_D = \frac{m_e}{1 + \frac{m_e}{m_d}}$$

Since $m_d > m_p$, it follows that $\frac{m_e}{m_d} < \frac{m_e}{m_p}$.

This implies that $1 + \frac{m_e}{m_d} < 1 + \frac{m_e}{m_p}$.

Since the denominator for $\mu_D$ is smaller than the denominator for $\mu_H$, and the numerator ($m_e$) is the same for both, it means:

$$\mu_D > \mu_H$$

Thus, the reduced mass of a deuterium atom is greater than the reduced mass of a hydrogen atom.