Chapter 28, Problem 42GP

(a)

To determine

The uncertainty in the energy of the $\text{n = 2}$ state.

Answer to Problem 42GP

  $\text{6}\text{.6}\text{×}{\text{10}}^{\text{-8}}\text{eV}$ .

Explanation of Solution

Given info:

Time is ${\text{10}}^{\text{-8}}\text{s}$ .

Formula used:

  $\text{ΔE }\ge \frac{\mathrm{\hslash }}{\text{Δt}}$

Where,

  $\mathrm{\hslash }$ is reduced Planck’s constant, $1.055\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$

  $\mathrm{\Delta }\text{E}$ is energy.

  $\mathrm{\Delta }\mathrm{t}$ is time.

Calculation:

Substitute the values in expression

  $\text{ΔE }\ge \frac{\mathrm{\hslash }}{\text{Δt}}$

  $\begin{array}{c}\mathrm{\Delta }\text{E=}\frac{\left(\text{1}\text{.055}\text{×}{\text{10}}^{\text{-34}}\text{J}\cdot \text{s}\right)}{{\text{10}}^{\text{-8}}\text{s}}\\ \text{=1}\text{.1}{\text{×10}}^{\text{-26}}\text{J}\\ \text{=}\text{6}\text{.6}{\text{×10}}^{\text{-8}}\text{eV}\end{array}$

Conclusion:

Because the ground state is stable, the uncertainty is associated with the excited state.

(b)

To determine

The fraction of the transition energy.

Answer to Problem 42GP

  $\text{6}\text{.5}{\text{×10}}^{\text{-9}}$

Explanation of Solution

Given info:

Time is ${\text{10}}^{\text{-8}}\text{s}$ .

Formula used:

Energy levels of $\text{H}$ atom is given by ${\text{E}}_{\text{n}}\text{=-}\frac{\text{13}\text{.6}}{{\text{n}}^{\text{2}}}\text{eV}$

Where, $\text{n}$ is the principal quantum number.

Calculation:

The transition energy is

  ${\text{E}}_{\text{n}}\text{=-}\left(\text{13}\text{.6 }\text{eV}\right)\left[\left(\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)\text{-}\left(\frac{\text{1}}{{\text{1}}^{\text{2}}}\right)\right]\text{=10}\text{.2}\text{eV}$

Now, $\frac{\text{ΔE}}{\text{E}}\text{=}\frac{\left(\text{6}\text{.6}\text{×}{\text{10}}^{\text{-8}}\text{eV}\right)}{\text{10}\text{.2}\text{eV}}\text{=}\text{6}\text{.5}{\text{×10}}^{\text{-9}}$

(c)

To determine

The wavelength and the width of the line.

Answer to Problem 42GP

  $\text{122}\text{nm}$

The width is $\text{7}\text{.9}{\text{×10}}^{\text{-7}}\text{nm}$ .

Explanation of Solution

Given info:

Time is ${\text{10}}^{\text{-8}}\text{s}$ .

The fraction of the transition energy is $\text{6}\text{.5}{\text{×10}}^{\text{-9}}$ .

Formula used:

  $\text{λ=}\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\text{E}}$

Where, $\text{λ}$ is wavelength in nm and $\text{E}$ is energy in eV.

Calculation:

The wavelength of the line is

  $\mathrm{\lambda }\text{=}\frac{\left(\text{1}\text{.24}\text{×}{\text{10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\left(\text{10}\text{.2}\text{eV}\right)}=1\text{22}\text{nm}$

Now calculate the width of the line by using the expression

  $\text{λ}+\mathrm{\Delta }\text{λ}\text{= }\frac{\left(\text{1}\text{.24}\text{×}{\text{10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\text{E+ΔE}}$

  $=\frac{\frac{\left(\text{1}\text{.24}\text{×}{\text{10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\text{E}}}{\left[\text{1+}\left(\frac{\text{ΔE}}{\text{E}}\right)\right]}\approx \text{λ}\left[\text{1-}\left(\frac{\text{ΔE}}{\text{E}}\right)\right]$

Ignore the sign, then

  $\begin{array}{c}\text{Δλ=λ}\left(\frac{\text{ΔE}}{\text{E}}\right)\\ \text{=}\left(\text{122}\text{nm}\right)\left(\text{6}{\text{.5×10}}^{\text{-9}}\right)\\ \text{=7}{\text{.9×10}}^{\text{-7}}\text{nm}\end{array}$