Sodium vapor lamps, use for public lighting, emits 489 nm yellow light. What energy is emitted by:
a) an excited sodium atom when it generates a photon?
b) 1 mole of excited sodium atoms at this wavelength?
a.
E = hc/wavelength
E = joules/photon
b.
answer in a x 6.02E23 = energy for a mole of photons in a.
what is ans this qustion
To determine the energy emitted by an excited sodium atom when it generates a photon, we can use the equation:
E = hc/λ
where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength in meters.
a) To find the energy emitted by an excited sodium atom at 489 nm, we need to convert the wavelength to meters first:
λ = 489 nm = 489 x 10^-9 m
Now we can use the equation:
E = (6.626 x 10^-34 J·s) x (2.998 x 10^8 m/s) / (489 x 10^-9 m)
Calculating this expression, we find:
E ≈ 4.04 x 10^-19 J
Therefore, an excited sodium atom emits approximately 4.04 x 10^-19 Joules of energy when it generates a photon.
b) To find the energy emitted by 1 mole of excited sodium atoms at this wavelength, we need to calculate the total energy by considering Avogadro's number (6.022 x 10^23) as the number of excited sodium atoms in 1 mole.
The energy emitted by 1 mole of excited sodium atoms can be obtained by multiplying the energy per atom by Avogadro's number:
Total energy = Energy per atom x Avogadro's number
Total energy = (4.04 x 10^-19 J) x (6.022 x 10^23 atoms)
Calculating this expression, we find:
Total energy ≈ 2.43 x 10^5 J
Therefore, 1 mole of excited sodium atoms emits approximately 2.43 x 10^5 Joules of energy at the wavelength of 489 nm.