What is the energy of photon whose wavelength is `6840 Å` ?

Options

A. `1.81 eV
B. 3.6 eV
C. `-13.6 eV
D. 12.1 eV

Explanation

To find the energy of a photon with a wavelength of \(6840 \, \text{Å}\), we can use the formula:

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

where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

### Step 1: Convert the wavelength from angstroms to meters

Given:
\[ \lambda = 6840 \, \text{Å} = 6840 \times 10^{-10} \, \text{m} = 6.840 \times 10^{-7} \, \text{m} \]

### Step 2: Substitute the values into the energy formula

Now we substitute the values of \(h\), \(c\), and \(\lambda\) into the energy formula:

\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{6.840 \times 10^{-7} \, \text{m}} \]

### Step 3: Calculate the energy in joules

Calculating the numerator:

\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \, \text{J m} \]

Now, divide by the wavelength:

\[ E = \frac{1.9878 \times 10^{-25}}{6.840 \times 10^{-7}} \approx 2.905 \times 10^{-19} \, \text{J} \]

### Step 4: Convert the energy from joules to electron volts

To convert joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):

\[ E \text{ (in eV)} = \frac{2.905 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 1.81 \, \text{eV} \]

### Final Answer

The energy of the photon whose wavelength is \(6840 \, \text{Å}\) is approximately \(1.81 \, \text{eV}\).

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