What is the energy of a photon in J emitted by the transition from the ni = 4 to the nf = 2 of a hydrogen atom?
E = 2.18E-18 J(1/2^2 - 1/4^2)
To determine the energy of a photon emitted during the transition from the initial energy level ni to the final energy level nf in a hydrogen atom, we can use the formula:
E = ΔE = hc/λ
Where:
- E is the energy of the photon (in joules)
- ΔE is the change in energy between the initial and final levels
- h is Planck's constant (h = 6.626 x 10^-34 J·s)
- c is the speed of light (c = 3.00 x 10^8 m/s)
- λ is the wavelength of the photon
Since we are given the initial energy level (ni = 4) and the final energy level (nf = 2) of the transition, we can calculate the change in energy ΔE using the formula:
ΔE = -Rh[(1/ni^2) - (1/nf^2)]
Where:
- Rh is the Rydberg constant for hydrogen (Rh = 2.18 x 10^-18 J)
Plugging in the values, we have:
ΔE = - (2.18 x 10^-18 J)[(1/4^2) - (1/2^2)]
= - (2.18 x 10^-18 J)(1/16 - 1/4)
= - (2.18 x 10^-18 J)(1/16 - 4/16)
= - (2.18 x 10^-18 J)(-3/16)
= (2.18 x 10^-18 J)(3/16)
= 3.27 x 10^-19 J
Now, using the energy formula E = ΔE = hc/λ, we can solve for the wavelength λ:
(2.18 x 10^-18 J)(3/16) = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s)/λ
Simplifying:
3.27 x 10^-19 J = (19.878 x 10^-26 J·m)/λ
To solve for λ, we rearrange the equation:
λ = (19.878 x 10^-26 J·m)/(3.27 x 10^-19 J)
≈ 6.064 x 10^-7 m
Therefore, the energy of a photon emitted during the transition from ni = 4 to nf = 2 in a hydrogen atom is approximately 6.064 x 10^-7 meters.