Chapter 27, Problem 76GP

To determine

The wavelength of light that would be emitted.

Answer to Problem 76GP

Wavelengths are $\text{653}\text{nm}$ , $\text{102}\text{nm}$ , $\text{122}\text{nm}$ .

Explanation of Solution

Given info:

The potential difference produces a kinetic energy of $12.3eV$ .

Formula used:

It is known that $\text{λ=}\frac{\text{hc}}{\left({\text{E}}_{\text{higher}\text{state}}{\text{-E}}_{\text{lower}\text{state}}\right)}$

Where,

  $\lambda$ is wavelength.

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

  $c$ is speed of light, ${\text{3×10}}^{\text{8}}\text{ m/s }$

  $\text{E}$ is energy of different transition states.

Calculation:

According to the question, it is possible to provide this much energy to hydrogen atom through collisions.

From the ground state, the maximum energy of the atom is $\text{-13}\text{.6}\text{eV}\text{+}\text{12}\text{.3}\text{eV}\text{=}\text{-1}\text{.3}\text{eV}$

From the energy level diagram, the atom could be excited to $n=3$ state. So, the possible transitions when the atom returns to ground state:

  $n=3$ to $n=2$

  $n=3$ to $n=1$

  $n=2$ to $n=1$

Now, calculation for wavelengths:

  ${\text{λ}}_{3\to 2}\text{=}\frac{\text{hc}}{\left({\text{E}}_{\text{3}}{\text{-E}}_{\text{2}}\right)}=\frac{\left(\text{1}{\text{.24×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\left[\text{-1}\text{.5}\text{eV-}\left(\text{-3}\text{.4}\text{eV}\right)\right]}=653\text{nm}$

  ${\text{λ}}_{3\to 1}\text{=}\frac{\text{hc}}{\left({\text{E}}_{\text{3}}{\text{-E}}_{\text{1}}\right)}=\frac{\left(\text{1}{\text{.24×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\left[\text{-1}\text{.5}\text{eV-}\left(\text{13}\text{.6}\text{eV}\right)\right]}=102\text{nm}$

  ${\text{λ}}_{2\to 1}\text{=}\frac{\text{hc}}{\left({\text{E}}_{\text{2}}{\text{-E}}_{\text{1}}\right)}=\frac{\left(\text{1}{\text{.24×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\left[\text{-3}\text{.4}\text{eV-}\left(\text{13}\text{.6}\text{eV}\right)\right]}=122\text{nm}$