The first member of the Balmer series of hydrogen atom has a wavelength of 6561 A. The wavelength of

Q: Why is the first member $n=3$ to $n=2$ and not $n=1$ to $n=2$?

Electrons emit photons when they fall from a *higher* energy level to a *lower* one. Since the Balmer series is defined by falling to $n=2$, the smallest possible fall must start from the very next level up, which is $n=3$.

Q: Do I need to memorize the value of the Rydberg constant $R$ for this?

Not at all! In ratio problems like this, $R$ is a common factor in both states. By dividing the two equations, $R$ elegantly cancels out, saving you from tedious calculations.

Q: How do I quickly convert between Angstroms and nanometers?

It's a simple factor of 10! $1 \text{ nm} = 10^{-9} \text{ m}$ and $1 \text{ \AA} = 10^{-10} \text{ m}$. Therefore, $1 \text{ nm} = 10 \text{ \AA}$. Just divide your Angstrom value by 10 to get nanometers.

Q: Why does the wavelength decrease for the second member?

Energy and wavelength are inversely proportional ($E = hc/\lambda$). The second member ($n=4 \to 2$) is a larger energy jump than the first member ($n=3 \to 2$). More energy means a shorter wavelength.

Q: What if the question asked for the series limit?

The 'series limit' refers to the largest possible jump, which is from $n=\infty$ down to $n=2$. You would just plug $n_2 = \infty$ into the Rydberg formula, making the $1/n_2^2$ term zero!

Question

The first member of the Balmer series of hydrogen atom has a wavelength of $6561 	ext{ \AA}$. The wavelength of the second member of the Balmer series (in nm) is ......... .

Sigma Percentile · Accepted Answer

### Decoding the Balmer Series

Imagine an electron inside a hydrogen atom as a person standing on a ladder. The rungs of this ladder represent the energy levels, denoted by the principal quantum number $n$. When the person jumps down to a lower rung, they release energy in the form of a photon.

The **Balmer series** is a specific set of jumps where the electron always lands on the second rung, $n=2$. The "first member" of this series is the shortest possible jump, which is from the very next rung, $n=3$, down to $n=2$. The question tells us that the photon emitted during this specific jump has a wavelength $\lambda_1 = 6561 	ext{ \AA}$.

### The Rydberg Formula: Our Master Key

To connect the energy levels to the wavelength of the emitted light, we use the elegant Rydberg formula:

$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$$

For the first member ($n_1 = 2$, $n_2 = 3$), we can substitute these values into our master key:

$$\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36}$$

### The Second Member: A Longer Jump

Now, let's look at the "second member" of the Balmer series. This corresponds to the next shortest jump, which is from $n=4$ down to $n=2$. Let's apply the Rydberg formula again to find its wavelength, $\lambda_2$:

$$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = \frac{3R}{16}$$

### The Art of Ratios

We now have two equations, but we don't know the exact value of the Rydberg constant $R$, and frankly, we don't need to! In physics, whenever you have two states of a system described by the same constant, taking a ratio is a powerful trick to simplify your life.

Let's divide the equation for $\lambda_1$ by the equation for $\lambda_2$:

$$\frac{\lambda_2}{\lambda_1} = \frac{5R/36}{3R/16}$$

The $R$ beautifully cancels out, leaving us with pure numbers:

$$\frac{\lambda_2}{\lambda_1} = \frac{5}{36} 	imes \frac{16}{3} = \frac{20}{27}$$

Now, we just plug in the known value of $\lambda_1 = 6561 	ext{ \AA}$:

$$\lambda_2 = 6561 	imes \frac{20}{27} = 243 	imes 20 = 4860 	ext{ \AA}$$

### The Final Trap: Units

We have our answer, $4860 	ext{ \AA}$, but we must be careful. The examiners have set a classic trap by asking for the answer in **nanometers (nm)**.

Recall the conversion factor: $1 	ext{ nm} = 10 	ext{ \AA}$. To convert our answer, we simply divide by 10:

$$\lambda_2 = \frac{4860}{10} 	ext{ nm} = 486 	ext{ nm}$$

And there we have it! The wavelength of the second member is **$486 	ext{ nm}$**.

Animated Solution for Physics - Atoms and Nuclei: The first member of the Balmer series of hydrogen atom has a wavelength of $6561 \overset{˚}{A}$ . The wavelength of the second member of the Balmer series (in nm) is ......... .

Enter Numerical Value:

Visualized Solution

\text{Balmer Series}

\text{First Member } (\lambda_1)

\text{Rydberg Formula for } \lambda_1

\text{Second Member } (\lambda_2)

\text{Rydberg Formula for } \lambda_2

\text{Ratio of Wavelengths}

\text{Calculating } \lambda_2

\text{Unit Conversion}

The Sigma Insight: Bohr's Atomic Model and Energy Levels

Decoding the Balmer Series

The Rydberg Formula

The Second Member

The Art of Ratios

The Final Trap

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