Animated Solution for Physics - Atoms and Nuclei: Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65\text{ A˚}$). The de-Broglie wavelength of this electron is

Imagine you are looking at a hydrogen atom. The electron isn't just sitting anywhere; it's in a very specific energy level. The problem tells us the electron is in the second excited state.

This is a classic trap where many students make a silly mistake! The lowest possible energy level is the ground state, which corresponds to the principal quantum number $n=1$. When the electron absorbs energy and jumps to the next level, it enters the first excited state ($n=2$). Therefore, jumping one more level puts it in the second excited state, meaning our principal quantum number is $n=3$.

Bohr Meets de Broglie

Niels Bohr proposed that an electron can only revolve in certain stable orbits where its angular momentum is quantized. Mathematically, Bohr's second postulate states:

Here, $m$ is the mass of the electron, $v$ is its velocity, $r_n$ is the radius of the $n^{\text{th}}$ orbit, and $h$ is Planck's constant.

Now, let's bring in Louis de Broglie. He revolutionized physics by suggesting that matter has wave-like properties. The de-Broglie wavelength $\lambda$ of a moving particle is given by:

Let's see what happens when we combine these two groundbreaking ideas. From Bohr's postulate, we can isolate the momentum $p=mv$:

Now, substitute this expression for momentum into de Broglie's wavelength equation:

Notice how beautifully Planck's constant $h$ cancels out! Rearranging the terms, we get a profound geometric relationship:

What does this mean physically? The term $2\pi r_n$ is simply the circumference of the electron's circular orbit. The equation tells us that the circumference must be exactly equal to an integer multiple of the electron's wavelength. In other words, the electron forms a perfect standing wave around the nucleus. If $n$ wasn't an integer, the wave would overlap out of phase and destructively interfere, destroying the orbit!

Now that we have our master equation, the rest is just plugging in the numbers. We know the electron is in the $n=3$ orbit, and the radius $r_3$ is given as $4.65\text{ A˚}$.

Substitute the values:

Looking at our options, the closest value is $9.7\text{ A˚}$. The elegance of this method is that we completely bypassed the need to calculate the electron's actual velocity, saving us time and preventing calculation errors!