The wavelength of th first spectral line of sodium 5896 Å . The fisrt excitation potential of sodium atomm will be (Planck's constant `h=6.63xx10^(-34) J-s)`

Options

A. `4.2 V`
B. `3.5 V`
C. `2.1 V`
D. None of these

Explanation

To find the first excitation potential of the sodium atom based on the given wavelength of its first spectral line, we can follow these steps:

### Step 1: Understand the relationship between energy, frequency, and wavelength
The energy (E) of a photon can be expressed in terms of its frequency (ν) using the equation:
\[ E = h \nu \]
Where:
- \( h \) is Planck's constant.

We can also express frequency in terms of wavelength (λ) using the speed of light (c):
\[ \nu = \frac{c}{\lambda} \]
Thus, we can rewrite the energy equation as:
\[ E = \frac{h c}{\lambda} \]

### Step 2: Substitute the known values
Given:
- Wavelength \( \lambda = 5896 \, \text{Å} = 5896 \times 10^{-10} \, \text{m} \)
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{J s} \)
- Speed of light \( c = 3 \times 10^{8} \, \text{m/s} \)

Substituting these values into the energy equation:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J s}) \times (3 \times 10^{8} \, \text{m/s})}{5896 \times 10^{-10} \, \text{m}} \]

### Step 3: Calculate the energy
Now, calculate the energy:
\[ E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^{8})}{5896 \times 10^{-10}} \]

Calculating the numerator:
\[ 6.63 \times 10^{-34} \times 3 \times 10^{8} = 1.989 \times 10^{-25} \, \text{J m} \]

Now, divide by the wavelength:
\[ E = \frac{1.989 \times 10^{-25}}{5896 \times 10^{-10}} \]

Calculating this gives:
\[ E \approx 3.37 \times 10^{-19} \, \text{J} \]

### Step 4: Convert energy to electron volts
To convert joules to electron volts, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]

Thus, converting the energy:
\[ E_{eV} = \frac{3.37 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.1 \, \text{eV} \]

### Step 5: Conclusion
The first excitation potential of the sodium atom is approximately:
\[ 2.1 \, \text{eV} \]