Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n=5 to n=1 in J.
To calculate the energy of a photon emitted during a transition in a hydrogen atom, we can use the formula:
E = -13.6 eV * (1/n_final^2 - 1/n_initial^2)
Where:
E is the energy of the photon emitted,
n_initial is the initial energy level of the electron (n=5),
n_final is the final energy level of the electron (n=1),
-13.6 eV is the energy of an electron in the first energy level of a hydrogen atom.
First, let's calculate the value within the parentheses:
(1/n_final^2 - 1/n_initial^2) = (1/1^2 - 1/5^2) = (1 - 1/25)
Simplifying further:
(1 - 1/25) = (25/25 - 1/25) = 24/25
Now, let's substitute the values into the formula:
E = -13.6 eV * (24/25)
To convert electron volts (eV) to Joules (J), we know that 1 eV is equal to 1.6 x 10^-19 J.
So, we can convert -13.6 eV to Joules as follows:
-13.6 eV * (1.6 x 10^-19 J/eV) = -2.176 x 10^-18 J
Finally, we can substitute this value into the formula:
E = (-2.176 x 10^-18 J) * (24/25)
Calculating this:
E = -2.067 x 10^-18 J
Therefore, the energy of the photon emitted during the transition from n=5 to n=1 in a hydrogen atom is -2.067 x 10^-18 J.
dE = 2.180E-18[1/(n1)^2 - (1/(n2)^2]
n1 = 1 and 1^2 = 1
n2 = 5 and 5^2 = 25