How to Calculate the Energy Released for an Electron in the Hydrogen Atom?
habitxo asked:
I’m extremely rusty in chemistry seeing as how I took Chem111 during my freshman year and am now a junior taking Chem112 and am having problems with this problem.
I’m extremely rusty in chemistry seeing as how I took Chem111 during my freshman year and am now a junior taking Chem112 and am having problems with this problem.
Calculate the energy released for an electron in the hydrogen atom that undergoes a transition from n = 2 to n = 4.
Multiple Choice Answers:
a. 4.09 X 10-19 J
b. 8.39 X 10-20 J
c. 7.46 X 10-32 J
d. 3.29 X 10-18 J
e. 5.39 X 10-21 J
Now, am I doing this right and using the correct formula?
I’m using the formula Ephoton = hv = hRH (1/n^2(low) – 1/n^2(high)) where hRH is 3.290×10^15 s^-1
so I get:
hv=3.290×10^15 s^-1 (1/4 – 1/16)
hv=3.290/0.1875
hv=17.54667
conversion to Joules:
17.54667 x 6.2415×10^18 = 1.095 x 10^20???
It’s all wrong so I threw it up there for someone to pick it apart and show me where I was wrong and if you could kindly show me the correct way, thanks!
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http://mooni.fccj.org/ ~ ethall / Rydberg / rydberg.htm onze vergelijkingen worden waarschijnlijk varianten van elkaar, maar ik graag gebruik maken van de Rydberg vergelijking in deze vorm de eerste plaats, en lossen voor golflengte: 1/wavelength = [1.0974x107m-1] [1/m2 - 1/n2] ====================== Vervolgens steek de stekker in: E = hc / golflengte, … op te lossen voor energie =================== maar laat ik het combineren van de twee vergelijkingen in een: E = ((6,625 e-34 J. Sec) (3 e8m/sec) [1.0974 e7m-1] [1/m2 - 1/n2] E = 2,18 e-18 [1/m2 - 1/n2] ============================ Nu laten we het oplossen van uw probleem: E = 2,18 e-18 [(1 / 4 - 1 / 16)] E = 2,18 e-18 (0,25 – 0,0625) E = 2,18 e-18 (0.1875) E = 4,0875 e-19 uw antwoord is a. 4,09 X 10-19 J