Chapter 28, Problem 52A

To determine

To calculate: The wavelength of laser needed to study the energy difference between $n=1$ and $n=2$ in hydrogen.

Answer to Problem 52A

The wavelength of laser is $\text{243 nm}$ .

Explanation of Solution

Given info:

Each photon has half the energy needed to excite the atoms from the ground state to the desired state.

The transition from $n=3$ to $n=2$ state.

Formula used:

The energy of $n$ th energy level of hydrogen is determined by

  $E_n=\frac{-13.6\text{ eV}}{{\text{n}}^2}$ .

Where $E_n$ represents the energy of the $\text{n}$ th energy level.

Energy of photon emitted:

  $E=\frac{hc}{\lambda }$

where $E$ is the energy of the photon, $h$ is the Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength of the photon.

Calculation:

As per problem,

Calculate the energy of the level $n=1$

  $\begin{array}{l}{E}_n=\frac{-13.6\text{ eV}}{{\text{n}}^2}\\ E_1=\frac{-13.6\text{ eV}}{{(1)}^2}\\ E_1=-13.6\text{ eV}\end{array}$

Calculate the energy of the level $n=2$

  $\begin{array}{l}{E}_n=\frac{-13.6\text{ eV}}{{\text{n}}^2}\\ E_2=\frac{-13.6\text{ eV}}{{(2)}^2}\\ E_2=-3.4\text{ eV}\end{array}$

The energy difference between the two levels is

  $\begin{array}{l}\Delta E=E_2-E_{\begin{array}{l}1\end{array}}\\ \Delta E=(-3.4\text{ eV})-(-13.6\text{ eV)}\\ \Delta E=10.2\text{ eV}\end{array}$

Each photon in the laser has half the excitation energy.

That is, energy of the photon $E_{photon}=\frac{\text{10}\text{.2 eV}}{2}=5.1\text{ eV}$

Therefore, wavelength of laser is given by

  $\begin{array}{l}{E}_{photon}=\frac{hc}{\lambda }\\ \lambda \text{=}\frac{hc}{E_{photon}}\end{array}$

Substitute the values, $c=3\times {10}^8\text{ m/s}$ , $h=6.63\times {10}^{-34}\text{ Js}$ and $E_{photon}=5.1\text{ eV}$

  $\begin{array}{l}\lambda \text{=}\frac{(6.63\times {10}^{-34}\text{ Js})(3\times {10}^8\text{ m/s})}{5.1\times 1.6\times {10}^{-19}\text{ J}}\text{ [multiply energy by }1.6\times {10}^{-19}\text{ to convert to joules]}\\ \lambda \text{=}2.43\times {10}^{-7}\text{ m}\\ \lambda \text{=243 nm}\end{array}$

Conclusion:

Thus, the wavelength of laser is $\text{243 nm}$ .