Question

The peak intensity of the CMBR occurs at a wavelength of 1.1mm. What is the energy in eV of a 1.1mm photon?

(a)1.13×10−⁶eV
(b)2.24×10−⁶eV
(c)3.48×10⁻⁶eV
(d)4.96×10⁻⁶eV

Answer 1

The energy of a 1.1 mm photon can be calculated using the equation E = hc / λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Plugging in the values, we find that the energy of a 1.1 mm photon is approximately 1.13 x 10^-6 eV.

Step-by-step explanation:

The energy of a photon can be calculated using the equation: E = hc / λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J*s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength. To convert the wavelength of 1.1 mm to meters, we divide it by 1000, giving 1.1 x 10^-3 m.

Plugging the values into the equation, we get: E = (6.63 x 10^-34 J*s * 3 x 10^8 m/s) / (1.1 x 10^-3 m) = 1.8 x 10^-20 J.

The energy in eV can be calculated by converting the energy from joules to electron volts using the conversion factor: 1 eV = 1.6 x 10^-19 J. So, the energy of a 1.1 mm photon is approximately 1.13 x 10^-6 eV.