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 a sodium atom given the wavelength of its first spectral line, we can use the relationship between energy and wavelength. The energy of a photon can be calculated using the formula:

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

Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

### Step 1: Convert Wavelength to Meters
The given wavelength is \( 5896 \, \text{Å} \) (angstroms). We need to convert this to meters:

\[ \lambda = 5896 \, \text{Å} = 5896 \times 10^{-10} \, \text{m} \]

### Step 2: Use the Energy Formula
Now, substitute the values into the energy formula:

\[ E = \frac{(6.63 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{5896 \times 10^{-10} \, \text{m}} \]

### Step 3: Calculate the Energy
Now, perform the calculation:

1. Calculate the numerator:
\[ 6.63 \times 10^{-34} \times 3.00 \times 10^8 = 1.989 \times 10^{-25} \, \text{J m} \]

2. Calculate the energy:
\[ E = \frac{1.989 \times 10^{-25}}{5896 \times 10^{-10}} = \frac{1.989 \times 10^{-25}}{5.896 \times 10^{-7}} \approx 3.375 \times 10^{-19} \, \text{J} \]

### Step 4: Convert Energy to Electron Volts
To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ E \approx \frac{3.375 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.11 \, \text{eV} \]

### Conclusion
The first excitation potential of the sodium atom is approximately \( 2.11 \, \text{eV} \).

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