Orders of magnitude 4: Assume again a Rubidium atom (mass m=87⋅u) is initially at rest. Again it then absorbs a photon of wavelength λ=780 nm. What is the kinetic energy Ekin of the atom after this process? Please give the energy in μK⋅kB units. This is an equivalent temperature unit: (Tequiv=E/kB), with the temperature in μK (within 1%).
To find the kinetic energy of the Rubidium atom after absorbing a photon, we can use the concept of conservation of energy. The energy of the photon is converted into the kinetic energy of the atom.
The energy of a photon can be calculated using the equation:
E = hc / λ
Where:
E is the energy of the photon
h is the Planck's constant (6.626 x 10^-34 J·s)
c is the speed of light (3 x 10^8 m/s)
λ is the wavelength of the photon
In this case, the wavelength (λ) is given as 780 nm. First, we need to convert it to meters by dividing by 1 billion:
λ = 780 nm / 1,000,000,000 = 7.8 x 10^-7 m
Now we can calculate the energy of the photon:
E = (6.626 x 10^-34 J·s * 3 x 10^8 m/s) / (7.8 x 10^-7 m)
E ≈ 8.06 x 10^-19 J
The kinetic energy of the atom can be found using the equation:
Ekin = E - E0
Where:
Ekin is the kinetic energy of the atom
E is the energy of the photon
E0 is the rest energy of the atom
The rest energy (E0) of the Rubidium atom can be calculated using the Einstein's mass-energy equation:
E0 = mc^2
Where:
m is the mass of the atom (87 u)
c is the speed of light (3 x 10^8 m/s)
First, we need to convert the mass of the atom from atomic mass units (u) to kilograms:
m = 87 u * (1.66 x 10^-27 kg / u)
m ≈ 1.44 x 10^-25 kg
Now we can calculate the rest energy of the atom:
E0 = (1.44 x 10^-25 kg) * (3 x 10^8 m/s)^2
E0 ≈ 1.29 x 10^-10 J
Finally, we can find the kinetic energy of the atom after the photon absorption:
Ekin = (8.06 x 10^-19 J) - (1.29 x 10^-10 J)
Ekin ≈ 7.93 x 10^-19 J
To convert this energy to μK·kB units, we need to use the conversion factor:
1 μK·kB = 1.38065 x 10^-29 J
Ekin_(μK·kB) = (7.93 x 10^-19 J) / (1.38065 x 10^-29 J)
Ekin_(μK·kB) ≈ 5.74 x 10^9 μK·kB
Therefore, the kinetic energy of the Rubidium atom after absorbing the photon is approximately 5.74 x 10^9 μK·kB (within 1% accuracy).