To find the energy associated with photons of wavelength 4000 Å (angstroms), we can use the formula for the energy of a photon:
\[ E = \frac{hc}{\lambda} \]
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 (\( 3 \times 10^{8} \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.
### Step 1: Convert the wavelength from angstroms to meters
1 angstrom (Å) = \( 10^{-10} \) meters, so:
\[ \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} = 4.0 \times 10^{-7} \, \text{m} \]
### Step 2: Substitute the values into the energy formula
Now we can substitute the values of \( h \), \( c \), and \( \lambda \) into the energy formula:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^{8} \, \text{m/s})}{4.0 \times 10^{-7} \, \text{m}} \]
### Step 3: Calculate the numerator
First, calculate the numerator:
\[ 6.63 \times 10^{-34} \times 3 \times 10^{8} = 19.89 \times 10^{-26} \, \text{J m} \]
### Step 4: Calculate the energy
Now, divide the numerator by the wavelength:
\[ E = \frac{19.89 \times 10^{-26}}{4.0 \times 10^{-7}} \]
Calculating this gives:
\[ E = 4.9725 \times 10^{-19} \, \text{J} \]
### Final Answer
Thus, the energy associated with photons of wavelength 4000 Å is approximately:
\[ E \approx 4.97 \times 10^{-19} \, \text{J} \]
---
