How to Calculate the Energy Released for an Electron in the Hydrogen Atom?

Question

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 3.290x10^15 s^-1

so I get:

hv=3.290x10^15 s^-1 (1/4 - 1/16)

hv=3.290/0.1875

hv=17.54667

conversion to Joules:

17.54667 x 6.2415x10^18 = 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!

Answer 1

http://mooni.fccj.org/~ethall/rydberg/ry...

our equations are likely to be variations of each other,

but I like to use the Rydberg equation in this form first, & solve for wavelength:

1/wavelength  = [1.0974x107m-1][1/m2 - 1/n2]

======================

http://cord.org/cm/leot/course01_mod05/m...

then plug it into : E = hc/ wavelength , ... to solve for energy

===================

but let me combine the two equations into one:

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]

============================

now let's solve your problem:

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

your answer is a. 4.09 X 10-19 J