Chapter 22, Problem 17Q

To determine

The energy of the photon when a hydrogen atom undergoes a spin-flip transition, and the number of protons that would be required to match the energy of a singlethe ${\text{H}}_{\text{α}}$ photon.

Answer to Problem 17Q

The energy of the photon emitted after the snip flip transition of the hydrogen atom is $9.5\times {10}^{-25}\text{J}$ and the photons required to carry the same energy as required by the ${\text{H}}_{\text{α}}$ photon is $3.2\times {10}^5$.

Explanation of Solution

Given:

The wavelength of the photon is, ${\lambda }_{{\text{H}}_{\alpha }}=656.3\text{nm}$.

The value of the plank’s constant is, $h=6.625\times {10}^{-34}\text{J}\cdot \text{s}$.

The wavelength of the photon after an hydrogen atom undergoes the snip flip transition is, $\lambda =21\text{cm}$.

Formula used:

The expression for the energy at the spin flip of transition of the photon is given by,

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

The expression for the energy of ${\text{H}}_{\text{α}}$ is given by,

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

The number of photons undergoing snip flip transition required for the same energy as $E_{{\text{H}}_{\text{α}}}$ is given by,

$n=\frac{E_{{\text{H}}_{\text{α}}}}{E_{\text{sp}}}$

Calculation:

The expression for the energy at the spin flip of transition of the photon is calculated as,

$\begin{array}{c}{E}_{\text{sp}}=\frac{hc}{\lambda }\\ =\frac{\left(6.625\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{21\text{cm}}\\ =\frac{\left(6.625\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(21\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =9.5\times {10}^{-25}\text{J}\end{array}$

The expression for the energy of ${\text{H}}_{\text{α}}$ iscalculated as,

$\begin{array}{c}{E}_{{\text{H}}_{\alpha }}=\frac{hc}{{\lambda }_{{\text{H}}_{\text{α}}}}\\ =\frac{\left(6.625\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{656.3\text{nm}}\\ =\frac{\left(6.625\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(656.3\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}\\ =3.0\times {10}^{-19}\text{J}\end{array}$

The number of photons undergoing snip flip transition required for the same energy as $E_{{\text{H}}_{\text{α}}}$ is calculated as,

$\begin{array}{c}n=\frac{E_{{\text{H}}_{\text{α}}}}{E_{\text{sp}}}\\ =\frac{3.0\times {10}^{-19}\text{J}}{9.5\times {10}^{-25}\text{J}}\\ =3.2\times {10}^5\end{array}$

Conclusion:

The energy of the photon emitted after the snip flip transition of the hydrogen atom is $9.5\times {10}^{-25}\text{J}$ and the photons required to carry the same energy as required by the ${\text{H}}_{\text{α}}$ photon is $3.2\times {10}^5$.