a.) Caculate the energy of light emitted when an electron moves from n=2 to n=1 in the hydrogen atom.
b.) Caculate the energy of light emitted when an electron moves from n=4 to n=2 in the hydrogen atom.
c.) How many total different light energies can be emitted from an electron moving down from n=5?
I don't know the equation for a.) and b.), and I don't understand c.). Help please!
E= 2.180E-18(1/n^2(1) - 1/n^2(2))
R = 1.09737E7
n1 = 1 so n^2 = 1
n2 = 2 so n^2 = 1/2^2 = 1/4
For b the only difference is
n1 = 2 so n^2 = 4
n2 = 4 so n^2 = 16
For c draw a diagram of energy levels that will look something like this.
n = 5 -----------------
n = 4 ----------------
n = 3 ---------------
n = 2 ---------------
n = 1 --------------
So an electron can move from
5 to 4;
5 to 3;
5 to 2;
5 to 1 and this results in 4 different energies being emitted.
After the 5 to 4, then the 4 can do the same thing; i.e.,
4 to 3;
4 to 2;
4 to 1; which results in 3 more energies being emitted.
Then the electron that goes from 5 to 3 can separately do
3 to 2; and
3 to 1; resulting in two more and finally, the one at 2 can do
2 to 1 resulting in 1 more. Add those together for the total. If I could draw a diagram on this site I could explain it much better.
Vry bad answers
a.) To calculate the energy of light emitted when an electron moves from n=2 to n=1 in the hydrogen atom, you can use the formula for the energy of a transition in the hydrogen atom:
E = -13.6 eV * (1/n_initial^2 - 1/n_final^2)
In this case, the initial energy level (n_initial) is 2, and the final energy level (n_final) is 1. Plugging these values into the formula, we get:
E = -13.6 eV * (1/2^2 - 1/1^2)
= -13.6 eV * (1/4 - 1/1)
= -13.6 eV * (1/4 - 1)
= -13.6 eV * (-3/4)
= 10.2 eV
So, the energy of light emitted when an electron moves from n=2 to n=1 in the hydrogen atom is 10.2 eV.
b.) Similarly, to calculate the energy of light emitted when an electron moves from n=4 to n=2 in the hydrogen atom, we use the same formula:
E = -13.6 eV * (1/n_initial^2 - 1/n_final^2)
In this case, the initial energy level (n_initial) is 4, and the final energy level (n_final) is 2. Plugging these values into the formula, we get:
E = -13.6 eV * (1/4^2 - 1/2^2)
= -13.6 eV * (1/16 - 1/4)
= -13.6 eV * (1/16 - 4/16)
= -13.6 eV * (-3/16)
= 2.55 eV
So, the energy of light emitted when an electron moves from n=4 to n=2 in the hydrogen atom is 2.55 eV.
c.) To understand how many different light energies can be emitted from an electron moving down from n=5, we need to consider the possible final energy levels (n_final) that the electron can transition to.
The energy of a transition in the hydrogen atom is given by the same formula as before:
E = -13.6 eV * (1/n_initial^2 - 1/n_final^2)
In this case, the initial energy level (n_initial) is 5. The final energy level (n_final) can be any positive integer smaller than n_initial, so we have n_final = 1, 2, 3, 4.
We can calculate the energy of light emitted for each transition using the formula, just like we did in parts a) and b). So, we have:
E_5to1 = -13.6 eV * (1/5^2 - 1/1^2) = -13.6 eV * (1/25 - 1) = 12.99 eV
E_5to2 = -13.6 eV * (1/5^2 - 1/2^2) = -13.6 eV * (1/25 - 1/4) = 10.20 eV
E_5to3 = -13.6 eV * (1/5^2 - 1/3^2) = -13.6 eV * (1/25 - 1/9) = 8.61 eV
E_5to4 = -13.6 eV * (1/5^2 - 1/4^2) = -13.6 eV * (1/25 - 1/16) = 7.74 eV
Therefore, there are four different light energies that can be emitted from an electron moving down from n=5 in the hydrogen atom: 12.99 eV, 10.20 eV, 8.61 eV, and 7.74 eV.