Animated Solution for Physics - Atoms and Nuclei: If the wavelength of the $n^{\text{th}}$ line of Lyman series is equal to the de-Broglie wavelength of electron in initial orbit of a hydrogen like element ($Z=11$). Find the value of $n$.

Have you ever wondered what happens when the macroscopic geometry of an electron's orbit perfectly matches the wavelength of the light it emits? This problem from JEE Advanced 2005 takes us on a thrilling journey where the de-Broglie wavelength of an electron in its orbit equals the wavelength of the photon it releases upon falling to the ground state. Let's dive into the cosmic dance of these wavelengths!

We are given a hydrogen-like element with an atomic number $Z=11$. The electron is initially in some excited state and makes a transition to the ground state, emitting a photon. This transition corresponds to the $n^{\text{th}}$ line of the Lyman series.

Since the Lyman series always ends at the ground state ($n_f=1$), the first line originates from the 2nd orbit, the second line from the 3rd orbit, and so on. Therefore, the $n^{\text{th}}$ line must originate from the $(n+1{)}^{\text{th}}$ orbit.

Our mission is to equate the wavelength of this emitted photon ($\lambda$) to the de-Broglie wavelength of the electron (${\lambda }_d$) in its initial orbit.

First, let's find the wavelength of the emitted photon using the Rydberg formula:

Substituting $n_f=1$, $n_i=n+1$, and $Z=11$, we get:

Next, we need the de-Broglie wavelength of the electron in its initial orbit. According to Bohr's quantization condition, the angular momentum is quantized:

Since the de-Broglie wavelength is ${\lambda }_d=\frac{h}{mv}$, we can rewrite the quantization condition as:

For our initial orbit $k=n+1$, the de-Broglie wavelength is:

The problem states that these two wavelengths are equal ($\lambda ={\lambda }_d$). Equating their reciprocals, we get:

We know that the radius of the $k^{\text{th}}$ orbit for a hydrogen-like atom is $r_k=\frac{k^2}{Z}{r}_0$. Substituting $r_{n+1}=\frac{(n+1{)}^2}{11}{r}_0$ into our equation:

Canceling the common terms from both sides, we arrive at a beautifully simplified equation:

Here comes the magic! The term $R(2\pi r_0)$ is a dimensionless constant. If we plug in the standard values for the Rydberg constant ($R\approx 1.097\times 10^7{\text{ m}}^{-1}$) and the Bohr radius ($r_0\approx 0.529\times 10^{-10}\text{ m}$), their product is approximately $\frac{1}{274}$. Remarkably, this is exactly equal to half the fine structure constant ($\frac{\alpha }{2}$).

Substituting this value back into our equation:

Multiplying both sides by 274 and rearranging the terms, we get a standard quadratic equation:

Solving this quadratic equation yields a positive root of $n\approx 24$. Since the principal quantum number must be an integer, the exact value is $n=24$.

This problem is a masterpiece that beautifully intertwines the macroscopic geometry of electron orbits with the quantum nature of emitted light!