Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition.
HOME > > Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition.

# Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition.

[From: ] [author: ] [Date: 12-06-25] [Hit: ]
h is Plancks constant, equal to 6.625 * 10**-34 joule-seconds,speed of light, 3.0 * 10**8 meters per second.......
The hydrogen atom undergoes a transition from n = 6 to n = 1.

-
If we can find the wavelength (w) we can find the energy (E) by: E = hc/w, where
h is Planck's constant, equal to 6.625 * 10**-34 joule-seconds, and c is the
speed of light, 3.0 * 10**8 meters per second. Since they are constants, their
product hc is constant, equal to 19.875 * 10**-26 joule-meters.
We can find the wavelength from the transitions by Rydberg's formula, which is
1/w = R(1/L² - 1/U²) where L and U are the lower and higher energy levels, given
as 1 & 6 in this example, and R is Rydberg's constant, experimentally determined
as 10967758 waves per meter for hydrogen. Substitute in known values and find
that 1/w = 10967758(1 - 1/36) = 10663097 so w = 9.378 * 10**-8 m = 93.78 nm.
The energy E = hc/w = (19.875 * 10**-26 J-m) / (9.378 * 10**-8 m.) Do the math
and we find that E = 2.119 * 10**-18 J, or 2.119 attojoules.