Suppose an atom absorbs light with a wavelength of 428 nm. How much energy did the atom gain?

To determine the energy gained by the atom when it absorbs light with a wavelength of \( \lambda = 428 \, \text{nm} \), we use the equation for the energy of a photon:

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

Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \),
- \( c \) is the speed of light, \( c = 3.00 \times 10^8 \, \text{m/s} \),
- \( \lambda \) is the wavelength of the light (in meters).

### Steps:
1. Convert the wavelength from nanometers to meters:
   \[
   \lambda = 428 \, \text{nm} = 428 \times 10^{-9} \, \text{m}
   \]
2. Plug the values into the equation:

\[
E = \frac{(6.626 \times 10^{-34}) \times (3.00 \times 10^8)}{428 \times 10^{-9}}
\]

Let me calculate that for you.The energy gained by the atom when it absorbs light with a wavelength of 428 nm is approximately \( 4.64 \times 10^{-19} \, \text{J} \).

Would you like further details or have any questions?

Here are 5 related questions to explore this concept further:
1. What is the relationship between wavelength and energy for photons?
2. How does the energy of a photon change if the wavelength decreases?
3. What is the energy in electronvolts (eV) for this same photon?
4. How is the absorption of light related to atomic excitation?
5. Can the energy of the absorbed light cause ionization of the atom?

**Tip:** The shorter the wavelength of light, the higher the energy of the photon!

Learn how to calculate the energy absorbed by an atom when it absorbs light with a wavelength of 428 nm. This problem uses the formula E = hc/λ, where Planck's constant and the speed of light are applied to find the energy. Suitable for high school students in Grades 10-12.

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