The approximate wavelength of a photon of energy `2.48 eV` is

Options

A. ` 500 Å `
B. `5000 Å `
C. `2000 Å `
D. `1000 Å `

Answer

To find the approximate wavelength of a photon with an energy of 2.48 eV, we can follow these steps:

### Step 1: Convert Energy from eV to Joules
The energy of the photon is given in electron volts (eV). We need to convert this energy into joules (J) using the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Thus, the energy in joules is:
\[ E = 2.48 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} \]
Calculating this gives:
\[ E = 2.48 \times 1.6 \times 10^{-19} \approx 3.968 \times 10^{-19} \text{ J} \]

### Step 2: Use the Energy-Wavelength Relation
The relationship between energy (E) and wavelength (λ) of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \text{ J s} \)
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \)

Rearranging this equation to solve for wavelength (λ):
\[ \lambda = \frac{hc}{E} \]

### Step 3: Substitute Values into the Equation
Now, we substitute the known values into the equation:
\[ \lambda = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{3.968 \times 10^{-19} \text{ J}} \]

### Step 4: Calculate the Wavelength
Calculating the numerator:
\[ 6.63 \times 10^{-34} \times 3 \times 10^8 \approx 1.989 \times 10^{-25} \text{ J m} \]
Now, divide by the energy:
\[ \lambda \approx \frac{1.989 \times 10^{-25}}{3.968 \times 10^{-19}} \approx 5.01 \times 10^{-7} \text{ m} \]
Converting this to nanometers (1 m = \( 10^9 \) nm):
\[ \lambda \approx 5.01 \times 10^{-7} \text{ m} = 501 \text{ nm} \]

### Step 5: Convert to Angstroms
Since \( 1 \text{ nm} = 10 \text{ Å} \):
\[ 501 \text{ nm} = 5010 \text{ Å} \]

### Conclusion
Thus, the approximate wavelength of the photon is:
\[ \lambda \approx 5000 \text{ Å} \]

### Final Answer
The correct option is approximately 5000 Å.

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