Question

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). For large 𝑛, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Answer 1

Step 1: Apply the formula for Energy of radiation at n level.
$v=\frac{m{e}^4}{(4\pi {)}^3{ϵ}_0^3{\left(\frac{h}{2\pi }\right)}^3}\times \left(\frac{1}{n^2}\right)$
The equation for energy $\left(E_1\right)$ of the radiation at the level n is
$E_1=h{v}_1$
$E_1=\frac{hm{e}^4}{(4\pi {)}^{3}epsilo{n}_0^2{\left(\frac{h}{2\pi }\right)}^3}\times \left(\frac{1}{n^2}\right).....\left(i\right)$
Here,
$v_1$ is the frequency at level n
h = Planck's constant
m = mass of the hydrogen atom
e = Electron charge
${ϵ}_0$ = Permittivity of free space
Step 2: Apply the formula for Energy of radiation at n -1 level.
The energy $\left(E_2\right)$ of the radiation at level (n -1) is
$E_2=h{v}_2$
$E_2=\frac{hm{e}^4}{(4\pi {)}^3{ϵ}_0^2{\left(\frac{h}{2\pi }\right)}^3}\times \left(\frac{1}{(n-1{)}^2}\right).....\left(ii\right)$
Here,
$v_2$ is the frequency at level (n -1)
Step 3: Calculate the difference in energies.
Energy (E) is released as a result of de-excitation
$E=E_2-E_1$
$hv=E_2-E_1.....\left(iii\right)$
Here,
v = Frequency of the radiation emitted
Substituting (i) and (ii) in (iii) we get
$v=\frac{m{e}^4}{(4\pi {)}^3{ϵ}_0^2{\left(\frac{h}{2\pi }\right)}^3}[\frac{1}{(n-1{)}^2}-\frac{1}{n^2}]$
$v=\frac{m{e}^4(2n-1)}{\left(4\pi {)}^3{ϵ}_0^2{\left(\frac{h}{2\pi }\right)}^3{n}^2\right(n-1{)}^2}$
For large n, we can write $(2n-1)\approx 2n$ and $(n-1)\approx n$
Therefore, $v=\frac{m{e}^4}{32{\pi }^3{ϵ}_0^2{\left(\frac{h}{2\pi }\right)}^3{\pi }^3}....\left(iv\right)$
Step 4: Find the value of frequency by putting the values obtained.
Classical relation of the frequency of revolution of an electron is given as
$v_c=\frac{v}{2\pi r}.....\left(v\right)$
In the $n^{th}$ orbit, the velocity of the electron is
$v=\frac{e^2}{4\pi {ϵ}_0\left(\frac{h}{2\pi }\right)n}.....\left(vi\right)$
The radius of the $n^{th}$ orbit r is given as
$r=\frac{4\pi {ϵ}_0{\left(\frac{h}{2\pi }\right)}^2}{m{e}^2}{n}^2......\left(vii\right)$
Putting the equation (vi) and equation (vii) in equation (v) we get
$v=\frac{m{e}^4}{32{\pi }^3{ϵ}_0^2{\left(\frac{h}{2\pi }\right)}^3{n}^3}....\left(viii\right)$
Therefore, the frequency of radiation emitted by the hydrogen atom is equal to the classical orbital frequency.
Final Answer: The frequency of radiation emitted by the hydrogen atom is equal to the classical orbital frequency.