1. Visible light has wavelengths in the range of \(400 \,\text{nm}\) to \(780 \,\text{nm}\). Calculate the range of energy of the photons of visible light.
Solution:
The energy (\(E\)) of a photon is calculated using the formula:
$$E = \frac{hc}{\lambda}$$
Given constants:
- Planck constant (\(h\)): \(6.63 \times 10^{-34} \,\text{J s}\)
- Speed of light (\(c\)): \(3 \times 10^8 \,\text{m s}^{-1}\)
- Wavelength range (\(\lambda\)): \(400 \,\text{nm}\) to \(780 \,\text{nm}\) (where \(1 \,\text{nm} = 10^{-9} \,\text{m}\))
1. Energy for the shortest wavelength (\(\lambda = 400 \,\text{nm}\)):
$$E_1 = \frac{(6.63 \times 10^{-34} \,\text{J s}) \times (3 \times 10^8 \,\text{m/s})}{400 \times 10^{-9} \,\text{m}}$$
$$E_1 = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} \,\text{J}$$
$$E_1 \approx 4.97 \times 10^{-19} \,\text{J}$$
2. Energy for the longest wavelength (\(\lambda = 780 \,\text{nm}\)):
$$E_2 = \frac{(6.63 \times 10^{-34} \,\text{J s}) \times (3 \times 10^8 \,\text{m/s})}{780 \times 10^{-9} \,\text{m}}$$
$$E_2 = \frac{19.89 \times 10^{-26}}{7.8 \times 10^{-7}} \,\text{J}$$
$$E_2 \approx 2.55 \times 10^{-19} \,\text{J}$$
2. Calculate the momentum of a photon of light of wavelength \(500 \,\text{nm}\).
Solution:
The momentum (\(p\)) of a photon is related to its wavelength (\(\lambda\)) by the de Broglie relation:
$$p = \frac{h}{\lambda}$$
Given values:
- Wavelength (\(\lambda\)): \(500 \,\text{nm} = 500 \times 10^{-9} \,\text{m}\)
- Planck constant (\(h\)): \(6.63 \times 10^{-34} \,\text{J s}\)
Calculation:
$$p = \frac{6.63 \times 10^{-34} \,\text{J s}}{500 \times 10^{-9} \,\text{m}}$$
$$p = \frac{6.63 \times 10^{-34}}{5 \times 10^{-7}} \,\text{kg m s}^{-1}$$
$$p = 1.326 \times 10^{-27} \,\text{kg m s}^{-1}$$
Final Answer:
The momentum of the photon is \(1.33 \times 10^{-27} \,\text{kg m s}^{-1}\).