Chapter 22, Problem 17Q

To determine

The energy of the photon emitted when a hydrogen atom undergoes a spin flip transition and find number of photons it would take to equal the energy of a single ${\text{H}}_{\text{α}}$ photon of wavelength 656.3 nm.

Answer to Problem 17Q

Solution:

$9.5\times {10}^{-25}\text{J,}\text{3}\text{.2}\times {\text{10}}^5$

Explanation of Solution

Given data:

The wavelength of ${\text{H}}_{\text{α}}$ photon is 656.3 nm.

Formula used:

The expression of energy of photons is:

$E=\frac{hc}{\lambda }$

Here, $h$ is Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength of the hydrogen atom.

The value of $h$ is $\text{6}{\text{.625×10}}^{\text{-34}}\text{J}\cdot \text{s}$ and $c$ is ${\text{3×10}}^{\text{8}}\text{m/s}$.

Explanation:

The wavelength of photon emitted after spin flip transition of an electron of hydrogen atom is $21\text{ cm}$.

Write the expression of energy at spin flip transition of photons:

$E_{sp}=\frac{hc}{\lambda }$

Substitute $\text{6}{\text{.625×10}}^{\text{-34}}\text{J}\cdot \text{s}$ for $h$, ${\text{3×10}}^{\text{8}}\text{m/s}$ for $c$ and $21\text{cm}$ for $\lambda$

$\begin{array}{c}{E}_{sp}=\frac{\left(\text{6}{\text{.625×10}}^{\text{-34}}\text{J}\cdot \text{s}\right)\left({\text{3×10}}^{\text{8}}\text{m/s}\right)}{\left(21\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ =9.5\times {10}^{-25}\text{J}\end{array}$

Write the expression of energy of ${\text{H}}_{\text{α}}$ photons:

$E_{H\alpha }=\frac{hc}{{\lambda }_{H\alpha }}$

Here, $E_{H\alpha }$ is the energy of ${\text{H}}_{\text{α}}$ photons and ${\lambda }_{H\alpha }$ is the wavelength of ${\text{H}}_{\text{α}}$ photons.

Substitute $\text{6}{\text{.625×10}}^{\text{-34}}\text{J}\cdot \text{s}$ for $h$, ${\text{3×10}}^{\text{8}}\text{m/s}$ for $c$ and $\text{656}\text{.3 nm}$ for $\lambda$.

$\begin{array}{c}{E}_{H\alpha }=\frac{\left(\text{6}{\text{.625×10}}^{\text{-34}}\text{J}\cdot \text{s}\right)\left({\text{3×10}}^{\text{8}}\text{m/s}\right)}{\left(656.3\text{nm}\left(\frac{1\text{m}}{{10}^{-9}\text{nm}}\right)\right)}\\ =3.0\times {10}^{-19}\text{J}\end{array}$

Determine the number of photons undergoing spin flip transition required for the same energy as $E_{H\alpha }$ as,

$\text{number of photons}=\frac{E_{sp}}{E_{H\alpha }}$

Substitute $9.5\times {10}^{-25}\text{J}$ for $E_{sp}$ and $3.0\times {10}^{-19}\text{J}$ for $E_{H\alpha }$:

$\begin{array}{c}\text{number of photons}=\frac{9.5\times {10}^{-25}\text{J}}{3.0\times {10}^{-19}\text{J}}\\ \text{=}\text{3}\text{.2}\times {\text{10}}^5\end{array}$

Conclusion:

Therefore, energy of the photon emitted when a hydrogen atom undergoes a spin flip transition is $9.5\times {10}^{-25}\text{J}$ and the number of photons required to carry the same energy as one ${\text{H}}_{\text{α}}$ photon is $\text{3}\text{.2}\times {\text{10}}^5$.