An electron in the n=7 level of the hydrogen atom relaxes to a lower energy level emitting light of 397 nm.

.............What is the value of "n" for the level to which the electron relaxed?...................

Ok, so this is what i know so far—we are suppose to use the energy formula. E=hc/ lambda

But, this is where i get confused. I do not know what else is suppose to be done. In some other yahoo answer, i saw some person use B( 1/n^2 - 1/n^2). Now that is where I got confused. No where in my Chem book did I see that formula. However, I did see this formula
Delta E = -2.18 * 10^-18 J (1/n^2f - 1/n^2f)

Where -2.18 * 10^ -18J is Rydberg constant for hydrogen

If this formula (the one with Rydberg constant) works to get the answer, can someone show me how this works with that formula? The answer should come up to "n=2"

we know the energy of the photon is given by

E = h c/L where h is planck's constant, c is the speed of light and L the wavelength, solving for E, we have

E = 6.63x10^-34Js 3x10^8m/s / 3.97x10^-7m = 5.01x10^-19 J

the energy levels in the H atom are described by

En = - 13.6eV/n^2 = - 2.18x10^-18J/n^2

so that the energy difference between two levels is

delta E = 2.18x 10^-18J(1/nl^2 - 1/nu^2) where nu is the upper energy level and nl is the lower

we know from the wavelength that the energy difference here is 5.01 x 10^-19J, so we have

5.01 x 10^-19J = 2.18x10^-18J(1/nl^2 - 1/7^2)

solve for nl:

0.23 = 1/nl^2 - 1/49

0.23 + 1/49 = 1/nl^2
nl = 2

Now that I look at this closer, I do not see how you got 0.23 from 5.01 x 10^-19 and 2.18 x 10^-18

Report Abuse