Determine the smallest wavelength in Lyman series among first four spectral lines in of hydrogen atom using Bohr's formula of energy quantisation .

Answer

To determine the smallest wavelength in the Lyman series among the first four spectral lines of the hydrogen atom using Bohr's formula of energy quantization, we can follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions where electrons fall to the first energy level (n=1) from higher energy levels (n=2, 3, 4, ...). The first four transitions in the Lyman series are:
1. n=2 to n=1
2. n=3 to n=1
3. n=4 to n=1
4. n=5 to n=1

### Step 2: Use Bohr's Formula for Energy Levels
The energy of an electron in the nth level of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

### Step 3: Calculate the Energy for Each Transition
1. **For n=2 to n=1:**
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]
\[ \Delta E_{2 \to 1} = E_1 - E_2 = -13.6 - (-3.4) = -10.2 \, \text{eV} \]

2. **For n=3 to n=1:**
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \, \text{eV} \]
\[ \Delta E_{3 \to 1} = E_1 - E_3 = -13.6 - (-1.51) = -12.09 \, \text{eV} \]

3. **For n=4 to n=1:**
\[ E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} \approx -0.85 \, \text{eV} \]
\[ \Delta E_{4 \to 1} = E_1 - E_4 = -13.6 - (-0.85) = -12.75 \, \text{eV} \]

4. **For n=5 to n=1:**
\[ E_5 = -\frac{13.6}{5^2} = -\frac{13.6}{25} \approx -0.544 \, \text{eV} \]
\[ \Delta E_{5 \to 1} = E_1 - E_5 = -13.6 - (-0.544) = -13.056 \, \text{eV} \]

### Step 4: Identify the Transition with the Highest Energy Difference
The transition with the highest energy difference will correspond to the smallest wavelength. From our calculations:
- \( \Delta E_{2 \to 1} = 10.2 \, \text{eV} \)
- \( \Delta E_{3 \to 1} = 12.09 \, \text{eV} \)
- \( \Delta E_{4 \to 1} = 12.75 \, \text{eV} \)
- \( \Delta E_{5 \to 1} = 13.056 \, \text{eV} \)

The transition from n=5 to n=1 has the highest energy difference.

### Step 5: Calculate the Wavelength Using the Energy Difference
Using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- Convert energy from eV to Joules: \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)

For \( \Delta E_{5 \to 1} = 13.056 \, \text{eV} \):
\[ \Delta E = 13.056 \times 1.6 \times 10^{-19} \, \text{J} \approx 2.089 \times 10^{-18} \, \text{J} \]

Now, substituting into the wavelength formula:
\[ \lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{2.089 \times 10^{-18}} \approx 9.5071 \times 10^{-8} \, \text{m} \approx 95.071 \, \text{nm} \]

### Final Answer
The smallest wavelength in the Lyman series among the first four spectral lines is approximately **95.071 nm**.

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