91影视

The energy of an orbit can be defined as the required energy to make the electron jump to infinity or the energy released when an electron from infinity is brought to that orbit.

As you know that energy of an orbit 鈥榥鈥� can be given by:

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\mathbf{-}\frac{\left|E_0\right|}{{\mathbf{n}}^{\mathbf{2}}}$

Where, $\left|E_0\right|$is the Energy of hydrogen鈥檚 ground state.

If the smallest possible jump at low-n end is $1鈫�2$, Its energy can be calculated as

$$\begin{gathered} {\mathrm{E}}_{1鈫�2}={\mathrm{E}}_2-{\mathrm{E}}_1 \\ =\frac{\left|{\mathrm{E}}_0\right|}{2^2}+\frac{\left|{\mathrm{E}}_0\right|}{1^2} \\ =\frac{3\left|{\mathrm{E}}_0\right|}{4} \end{gathered}$$

Now, let n and n-1 are the orbits for jump in the high-n end, then their energy difference can be calculated as

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}-1鈫�\mathrm{n}}={\mathrm{E}}_{\mathrm{n}}-{\mathrm{E}}_{\mathrm{n}-1} \\ =-\frac{\left|{\mathrm{E}}_0\right|}{{\mathrm{n}}^2}+\frac{\left|{\mathrm{E}}_0\right|}{{\left(\mathrm{n}-1\right)}^2} \\ =\left|{\mathrm{E}}_0\right|\frac{2\mathrm{n}-1}{{\mathrm{n}}^2{\left(\mathrm{n}-1\right)}^2} \end{gathered}$$

If n is high, you can write the above expression as $\frac{\left|{\mathrm{E}}_0\right|}{{\mathrm{n}}^3}$.

Hence, the photon energy released in the smallest possible jumps on low and high-n sides are $3\left|{\mathrm{E}}_0\right|/4$and $2\left|{\mathrm{E}}_0\right|/{\mathrm{n}}^3$respectively, where ${\mathrm{E}}_0$is hydrogen鈥檚 ground-state energy.

Given,

F=ma 鈥�.. (1)

As you know that Coulomb force between two charges separated by a distance r is given as,

$\mathrm{F}=\frac{{\mathrm{e}}^2}{4{\mathrm{蟿蟿蔚}}_0{\mathrm{r}}^2}$ 鈥�.. (2)

You also know that acceleration of the charge can be written as

$\mathrm{a}={\mathrm{蝇}}^2\mathrm{r}$ 鈥�.. (3)

Now, by using equations (2) and (3) in equation (1) you get,

$\frac{{\mathrm{e}}^2}{4{\mathrm{蟿蟿蔚}}_0{\mathrm{r}}^2}={\mathrm{m蝇}}^2\mathrm{r}$

$\mathrm{蝇}=\sqrt{\frac{{\mathrm{e}}^2}{4{\mathrm{蟿蟿蔚}}_0{\mathrm{mr}}^3}}$ 鈥�.. (4)

Hence, the angular velocity of point charge is $\mathrm{蝇}=\sqrt{\frac{{\mathrm{e}}^2}{4{\mathrm{蟿蟿蔚}}_0{\mathrm{mr}}^3}}$.

Consider the given equation for angular frequency as below.

$\mathrm{r}={\mathrm{a}}_0{\mathrm{n}}^2$ 鈥�.. (5)

Now, by using eq. (5) in eq. (4), you get,

$\mathrm{蝇}=\sqrt{\frac{{\mathrm{e}}^2}{4{\mathrm{蟿蟿蔚}}_0{\mathrm{ma}}_0{}^3{\mathrm{n}}^6}}$

If,${\mathrm{a}}_0=4{\mathrm{蟿蟿蔚}}_0{\mathrm{h}}^2/{\mathrm{me}}^2$

$$\begin{gathered} \mathrm{蠄}=\sqrt{\frac{{\mathrm{e}}^2}{4{\mathrm{蟿蟿蔚}}_0{\left(4{\mathrm{蟿蟿蔚}}_0{\mathrm{h}}^2/{\mathrm{me}}^2\right)}^3{\mathrm{n}}^6}} \\ =\frac{{\mathrm{me}}^4}{2{\left(4{\mathrm{蟿蟿蔚}}_0\right)}^2{\mathrm{h}}^2}\frac{2}{\mathrm{h}}\frac{1}{\mathrm{n}} \\ =2\frac{\left|{\mathrm{E}}_0\right|}{\mathrm{h}}\frac{1}{{\mathrm{n}}^3} \end{gathered}$$

Hence, the angular frequency of photon $2\frac{\left|{\mathrm{E}}_0\right|}{\mathrm{h}}\frac{1}{{\mathrm{n}}^3}$which matches with the high-n end side.