91影视

Energy levels or energy state is any discrete value from a set of total energy values for a force-limited subatomic particle in a limited space, or for a system of such particles, such as an atom or a nucleus.

According to the law of conservation of momentum, the total momentum of two isolated bodies remains same unless an external force is applied.

As you know that, energy difference,

${\mathrm{E}}_{\mathrm{n}}=13.6\mathrm{eV}\left(\frac{1}{{{\mathrm{n}}_{\mathrm{f}}}^2}-\frac{1}{{{\mathrm{n}}_{\mathrm{i}}}^2}\right)$

Where, ${\mathrm{n}}_{\mathrm{f}}$is the final orbit and ${\mathrm{n}}_{\mathrm{i}}$is the initial orbit.

Now, energy difference for transition from n = 2 to n = 1 is

role="math" localid="1659178231672" $$\begin{gathered} 鈭�\mathrm{E}={\mathrm{E}}_1-{\mathrm{E}}_2 \\ =\left(13.6\mathrm{eV}\right)\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right) \\ =\left(13.6\mathrm{eV}\right)\left(1-0.25\right) \\ =10.2\mathrm{eV} \\ 鈭�\mathrm{E}=10.2\mathrm{eV}脳1.6脳{10}^{-19}\mathrm{J}/\mathrm{eV} \\ =1.63脳{10}^{-18}\mathrm{J} \end{gathered}$$

As you know that, Energy of emitted photon,

$E=\frac{hc}{位}$

Where, h is Plank鈥檚 Constant, c is the speed of light, and $位$is the wavelength.

The relation between wavelength and momentum is as below.

$\frac{\mathrm{hc}}{\mathrm{位}}=\mathrm{pc}$

Where, p is the momentum.

Therefore,

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{E}}{\mathrm{c}} \\ =\frac{1.63脳{10}^{-18}}{3脳{10}^8\mathrm{m}/\mathrm{s}} \\ =5.44脳{10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

As you know that,

$\mathrm{Momentum}\left(\mathrm{p}\right)=\mathrm{mass}\left(\mathrm{m}\right)脳\mathrm{Velocity}\left(\mathrm{v}\right)$

Also, the kinetic energy is,

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

Hence, by comparing both the above equation, you have

$\mathrm{KE}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}$

Since, momentum must be conserved, the atom will have an equal and opposite momentum to balance that out.

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{KE}=\frac{{(5.44脳{10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s})}^2}{2脳1.67脳{10}^{-27}\mathrm{kg}} \\ =8.9脳{10}^{-27}\mathrm{J} \\ =5.5脳{10}^{-8}\mathrm{eV} \end{gathered}$$

Hence, the recoil energy is $\mathrm{KE}=5.5脳{10}^{-8}\mathrm{eV}$.

From Step 1 you, have,

Photon鈥檚 Energy, E = 10.2 eV

Recoil energy, $\mathrm{KE}=5.5脳{10}^{-8}\mathrm{eV}$

Take a ratio of the recoil energy and photon鈥檚 energy as below.

$$\begin{gathered} \frac{\mathrm{Recoil}\mathrm{Energy}}{\mathrm{Photon}\text{'}\mathrm{s}\mathrm{Energy}}=\frac{5.5脳{10}^{-8}\mathrm{eV}}{10.2\mathrm{eV}} \\ =5脳{10}^{-9} \end{gathered}$$

Hence, the recoil energy is the $5脳{10}^{-9}$of the photon鈥檚 energy.