Animated Solution for Chemistry - Atomic Structure: Energy of an electron is given by $E=-2.178\times 10^{-18}\text{ J}\left(\frac{Z^2}{n^2}\right)$. Wavelength of light required to excite an electron in an hydrogen atom from level $n=1$ to $n=2$ will be ($h=6.62\times 10^{-34}\text{ J s}$ and $c=3.0\times 10^8{\text{ ms}}^{-1}$)

Exciting a Hydrogen Atom

Imagine you are looking at a single hydrogen atom. Deep within its structure, an electron is sitting comfortably in its lowest possible energy state, known as the ground state ($n=1$). Our mission in this problem is to excite this electron, forcing it to make a quantum leap to the first excited state ($n=2$).

But electrons don't just jump on their own; they need a precise packet of energy to make the transition. This packet of energy is delivered by a photon of light. The question asks us to find the exact wavelength of this photon.

The problem generously provides us with the master formula for the energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like species:

Notice the negative sign? That is a crucial piece of physics. It tells us that the electron is bound to the nucleus. You would have to supply energy to free it completely (bringing its energy to zero at $n=\infty$). For a hydrogen atom, the atomic number $Z$ is simply $1$.

To make the leap from $n=1$ to $n=2$, the electron must absorb a photon whose energy exactly matches the energy difference between these two levels. Let's set up the equation for this energy difference, $\Delta E$:

Substituting our formula, we get:

Now, let's do the math. Inside the parentheses, $1-\frac{1}{4}$ gives us $\frac{3}{4}$. Multiplying this fraction by our energy constant:

This is the exact amount of energy the incoming photon must possess.

We have the energy, but the question asks for the wavelength. This is where the famous Planck-Einstein relation comes into play:

Here, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. Let's rearrange this equation to solve for $\lambda$:

Now, we carefully substitute the given values. Don't rush through this part; silly mistakes with powers of 10 are common here.

Solving the numerator gives us $19.86\times 10^{-26}$. Dividing this by our denominator:

Adjusting the decimal point to match standard scientific notation, we get:

This perfectly matches option (a). What a beautiful and precise application of quantum mechanics!