Calculate the wavelength of the spectral line when an electron jumps from n=7 to n=4 level in an atom of hydrogen.

Answer

To calculate the wavelength of the spectral line when an electron jumps from the n=7 to n=4 level in a hydrogen atom, we will follow these steps:

### Step 1: Determine the formula for change in energy (ΔE)
The change in energy when an electron transitions between energy levels in a hydrogen atom is given by the formula:

\[ \Delta E = 2.2 \times 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( n_1 \) is the lower energy level and \( n_2 \) is the higher energy level.

### Step 2: Assign values to \( n_1 \) and \( n_2 \)
In this case, since the electron is jumping from \( n=7 \) to \( n=4 \):
- \( n_1 = 4 \)
- \( n_2 = 7 \)

### Step 3: Substitute the values into the energy formula
Now, we substitute \( n_1 \) and \( n_2 \) into the energy formula:

\[ \Delta E = 2.2 \times 10^{-18} \left( \frac{1}{4^2} - \frac{1}{7^2} \right) \]

Calculating the squares:
- \( 4^2 = 16 \)
- \( 7^2 = 49 \)

Now substituting these values:

\[ \Delta E = 2.2 \times 10^{-18} \left( \frac{1}{16} - \frac{1}{49} \right) \]

### Step 4: Calculate the fractions
Calculating \( \frac{1}{16} \) and \( \frac{1}{49} \):

\[ \frac{1}{16} = 0.0625 \]
\[ \frac{1}{49} \approx 0.020408 \]

Now, subtract these two values:

\[ \frac{1}{16} - \frac{1}{49} \approx 0.0625 - 0.020408 = 0.042092 \]

### Step 5: Calculate ΔE
Now substitute this back into the ΔE formula:

\[ \Delta E = 2.2 \times 10^{-18} \times 0.042092 \approx 9.27 \times 10^{-20} \text{ Joules} \]

### Step 6: Relate energy to wavelength
The energy emitted during the transition can also be expressed in terms of wavelength using the equation:

\[ \Delta E = \frac{hc}{\lambda} \]

where:
- \( h = 6.626 \times 10^{-34} \) J·s (Planck's constant)
- \( c = 3 \times 10^8 \) m/s (speed of light)

### Step 7: Rearrange to find wavelength (λ)
Rearranging the equation to solve for \( \lambda \):

\[ \lambda = \frac{hc}{\Delta E} \]

### Step 8: Substitute values into the wavelength formula
Now substitute the values of \( h \), \( c \), and \( \Delta E \):

\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{9.27 \times 10^{-20}} \]

Calculating the numerator:

\[ 6.626 \times 10^{-34} \times 3 \times 10^8 \approx 1.9878 \times 10^{-25} \]

Now divide by \( \Delta E \):

\[ \lambda \approx \frac{1.9878 \times 10^{-25}}{9.27 \times 10^{-20}} \approx 2.14 \times 10^{-7} \text{ meters} \]

### Step 9: Convert to Angstroms
To convert meters to Angstroms (1 Angstrom = \( 10^{-10} \) meters):

\[ \lambda \approx 2.14 \times 10^{-7} \text{ m} = 2140 \text{ Angstroms} \]

### Final Answer
The wavelength of the spectral line when an electron jumps from n=7 to n=4 level in a hydrogen atom is approximately **2140 Angstroms**.

---