The energy of a photon (in ev) of wavelength 5000 Å will be

Options

A. 2.48 ev
B. 8.42 ev
C. zero
D. 4.82 ev

Answer

To find the energy of a photon with a wavelength of 5000 Å (angstroms), we can use the formula derived from Planck's quantum theory:

### Step-by-Step Solution:

**Step 1: Understand the formula for energy of a photon.**
The energy (E) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy in joules,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

**Step 2: Convert the wavelength from angstroms to meters.**
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \), we convert 5000 Å to meters:
\[ \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

**Step 3: Substitute the values into the energy formula.**
Now, substitute \( h \), \( c \), and \( \lambda \) into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{5 \times 10^{-7} \, \text{m}} \]

**Step 4: Calculate the energy in joules.**
Calculating the numerator:
\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \, \text{Jm} \]
Now, divide by the wavelength in meters:
\[ E = \frac{1.9878 \times 10^{-25}}{5 \times 10^{-7}} = 3.9756 \times 10^{-19} \, \text{J} \]

**Step 5: Convert joules to electron volts.**
To convert joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{3.9756 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.485 \, \text{eV} \]

**Step 6: Round the result.**
Rounding to two decimal places, we find:
\[ E \approx 2.48 \, \text{eV} \]

### Final Answer:
The energy of a photon with a wavelength of 5000 Å is approximately **2.48 eV**.

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