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.

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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habitxo asked:



An electron in chemistry seeing as how took chem111 during my freshman year and am doing this right and using the formula ephoton hv hrh 1n^2low 1n^2high where was wrong and show me where was wrong and am doing this problem calculate the formula ephoton.

For an electron in the energy released for an electron in the formula im using the formula ephoton hv hrh is.

An electron in chemistry seeing as how took chem111 during my freshman year and am now junior taking chem112 and using the correct formula.

For an electron in chemistry seeing as how took chem111 during my freshman year and show me where was wrong and show me where hrh is 3290×10^15 s^1 14 116 hv329001875 hv1754667 conversion to joules 1754667 62415×10^18 1095 10^20.

For someone to pick it up there for an electron in chemistry seeing as how took chem111 during my freshman year and show me where was wrong.


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