Find the longest wavelength present in the Balmer series of hydrogen.

Explanation

To find the longest wavelength present in the Balmer series of hydrogen, we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The longest wavelength in this series corresponds to the transition with the smallest energy difference.

### Step 2: Identify the Relevant Energy Levels
For the Balmer series, the relevant transitions are:
- n = 3 to n = 2
- n = 4 to n = 2
- n = 5 to n = 2
- etc.

The transition with the smallest energy difference will be from n = 3 to n = 2.

### Step 3: Calculate the Energy Difference
The energy levels of the hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

For the transition from n = 3 to n = 2:
- Energy at n = 3:
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV} \]

- Energy at n = 2:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

Now, find the energy difference (ΔE):
\[ \Delta E = E_2 - E_3 = (-3.4 \, \text{eV}) - (-1.51 \, \text{eV}) = -3.4 + 1.51 = -1.89 \, \text{eV} \]

### Step 4: Calculate the Wavelength
The wavelength (λ) can be calculated using the formula:
\[ \lambda = \frac{hc}{E} \]
Where:
- \( h \) = Planck's constant = \( 4.1357 \times 10^{-15} \, \text{eV s} \)
- \( c \) = speed of light = \( 3 \times 10^8 \, \text{m/s} \)

Using the value of \( E = 1.89 \, \text{eV} \):
\[ \lambda = \frac{12375 \, \text{eV nm}}{1.89 \, \text{eV}} \]

Calculating this gives:
\[ \lambda \approx 6546.02 \, \text{nm} \]

### Step 5: Convert to Angstroms
Since 1 nm = 10 Å:
\[ \lambda \approx 6546.02 \, \text{nm} = 65460.2 \, \text{Å} \]

### Step 6: Final Answer
The longest wavelength present in the Balmer series of hydrogen is approximately:
\[ \lambda \approx 651 \, \text{nm} \]