Chapter 27, Problem 74GP

To determine

The fraction by which $\text{H}$ atom decreases its mass and makes $\text{n}=\text{3}\text{to}\text{n}=\text{1}$ transition.

Answer to Problem 74GP

  $-1.29\times {10}^{-8}$

Explanation of Solution

Given data:

Upper state, $n=3$

Lower state, $n=1$

Formula used:

  $\frac{\Delta m}{m_0}=\frac{\Delta E}{m_0{c}^2}$

Where,

  $\Delta E$ is change in energy.

  $m_0{c}^2$ is minimum photo energy.

  $m_0$ is initial mass of atom.

  $\Delta m$ is change in mass due to emission of photon.

Calculation:

The decrement of mass in atom is due to photon emission.

Now, substitute the data into the formula

  $\frac{\Delta m}{m_0}=\frac{\Delta E}{m_0{c}^2}$

  $\therefore \frac{\Delta m}{m_0}=\frac{\left(-13.6eV\right)\left[\left(\frac{1}{1^2}\right)-\left(\frac{1}{3^2}\right)\right]}{\left(939\times {10}^{6}eV\right)}$

  $\therefore \frac{\Delta m}{m_0}=-1.29\times {10}^{-8}$

Conclusion: The fraction of mass decreases by hydrogen atom is $-1.29\times {10}^{-8}$ .