Question

An electron in a hydrogen atom in the n = 6 energy level emits 109.4 kJ/mol of energy in a transition to a lower energy level.
To what energy level does the electron fall?

Answer 1

First question, nice

Answer 2

hi, do you still need help?

Answer 3

Yes

Answer 4

I've tried Rydberg's equation and also Bohr's but no idea

\(\Large \frac{1}{\lambda}=R\left(\frac{1}{n^{2}_{f}} - \frac{1}{n^{2}_{i}}\right)\)

After using
\(\Large \Delta E = \frac{hc}{\lambda}\)

And the other formula

\(\Large \Delta E = -b\left(\frac{1}{n^{2}_{f}} - \frac{1}{n^{2}_{i}}\right)\)

where
b = 2.178 * 10^-18 J
h = 6.63 * 10^-34 J s
c = 2.998 * 10^8 m s^-1
R = 1.0974 x 107 m^-1

Answer 5

\(\Large \Delta E = -b\left(\frac{1}{n^{2}_{f}} - \frac{1}{n^{2}_{i}}\right)\)

\( 109.4 ~kJ~mol^{-1}= -2.178\cdot 10^{-18} J\left(\frac{1}{n^{2}_{f}} - \frac{1}{6^2}\right)\)

\(\Large \frac{109400 ~J~mol^{-1}}{ -2.178\cdot 10^{-18} J}=\left(\frac{1}{n^{2}_{f}} - \frac{1}{6^2}\right)\)

\(\Large -5.02295684e22~ mol^{-1} + \frac{1}{36}=\frac{1}{n^{2}_{f}} \)

like idek what I'm doing x'd

Answer 6

The formula for the energy is E(n) = -13.6 eV * (1/n^2)
E(n) - E(6) = 13.6 eV * (1/n^2 - 1/6^2) = 109.4 kJ /mol

109.4 kJ / mol = 109.4 * (10^3 J / kJ) * (1 eV / 1.6 * 10^-19 J) * (1 mol / 6.02 *10^23 e) * (1 e / atom)
= 1.14 eV

n = [1.14 eV / 13.6 eV + 1/6^2]^(-1/2) = 2.99... ~ 3.

The answer is n = 3.

Answer 7

Dang that was long! LOL cxxx

Answer 8

LowKey, I don't need a copy pasted answer from some other source. Can you explain where you got the formula E(n) = -13.6 eV * (1/n^2) from?

Answer 9

Well niels Bohr derived the formula for hydrogen like species \[E(n)=-13.6 \frac{z^2}{n^2}eV\]