Chapter 37, Problem 24P

(a)

To determine

The reduced mass of a diatomic molecule is always smaller than the mass of the molecule and to calculate the reduced mass for ${\text{H}}_2$ .

Answer to Problem 24P

Thereduced mass of ${\text{H}}_2$ is $0.500\text{u}$ .

Explanation of Solution

Formula used:

The expression for reduced mass of a two body system is given by,

  $\mu =\frac{m_1{m}_2}{m_1+m_2}$

Calculation:

Dividing the expression for reduced mass of a two body systemby $m_2$ ,

  $\begin{array}{c}\mu =\frac{m_1{m}_2}{m_1+m_2}\\ =\frac{m_1}{1+\frac{m_1}{m_2}}\end{array}$

Again, dividing the expression for reduced mass of a two body system by $m_1$

  $\begin{array}{c}\mu =\frac{m_1{m}_2}{m_1+m_2}\\ =\frac{m_2}{1+\frac{m_2}{m_1}}\end{array}$

In both case, the denominator is greater than 1 which implies

  $\mu <m_1$ and $\mu <m_2$

The reduced mass of ${\text{H}}_2$ is calculated as,

  $\begin{array}{c}\mu =\frac{m_1{m}_2}{m_1+m_2}\\ =\frac{\left(1\text{u}\right)\left(1\text{u}\right)}{\left(1\text{u}\right)+\left(1\text{u}\right)}\\ =0.500\text{u}\end{array}$

Conclusion:

Therefore, thereduced mass of ${\text{H}}_2$ is $0.500\text{u}$ .

(b)

To determine

The reduced mass of ${\text{N}}_2$ .

Answer to Problem 24P

The reduced mass of ${\text{N}}_2$ is $7.00\text{u}$ .

Explanation of Solution

Calculation:

The reduced mass of ${\text{N}}_2$ is calculated as,

  $\begin{array}{c}\mu =\frac{m_1{m}_2}{m_1+m_2}\\ =\frac{\left(14\text{u}\right)\left(14\text{u}\right)}{\left(14\text{u}\right)+\left(14\text{u}\right)}\\ =7.00\text{u}\end{array}$

Conclusion:

Therefore, the reduced mass of ${\text{N}}_2$ is $7.00\text{u}$ .

(c)

To determine

The reduced mass of $\text{CO}$ .

Answer to Problem 24P

The reduced mass of $\text{CO}$ is $6.86\text{u}$ .

Explanation of Solution

Calculation:

The reduced mass of $\text{CO}$ is calculated as,

  $\begin{array}{c}\mu =\frac{m_1{m}_2}{m_1+m_2}\\ =\frac{\left(12\text{u}\right)\left(16\text{u}\right)}{\left(12\text{u}\right)+\left(16\text{u}\right)}\\ =6.86\text{u}\end{array}$

Conclusion:

Therefore, the reduced mass of $\text{CO}$ is $6.86\text{u}$ .

(d)

To determine

The reduced mass of $\text{HCl}$ .

Answer to Problem 24P

The reduced mass of $\text{HCl}$ is $0.973\text{u}$ .

Explanation of Solution

Calculation:

The reduced mass of $\text{HCl}$ is calculated as,

  $\begin{array}{c}\mu =\frac{m_1{m}_2}{m_1+m_2}\\ =\frac{\left(1\text{u}\right)\left(\text{35}\text{.5}\text{u}\right)}{\left(1\text{u}\right)+\left(\text{35}\text{.5}\text{u}\right)}\\ =0.973\text{u}\end{array}$

Conclusion:

Therefore, the reduced mass of $\text{HCl}$ is $0.973\text{u}$ .