## Test 2 #5

$E=hv$

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PranithaPrasad
Posts: 58
Joined: Fri Sep 29, 2017 7:06 am
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### Test 2 #5

5. Determine the frequency of the emitted photon when an electron in a hydrogen atom drops from the quantum state of n=3 to n=1.

The answer is apparently 2.92 x 10^15. Could someone explain how we would get that?

Curtis Tam 1J
Posts: 105
Joined: Thu Jul 13, 2017 3:00 am

### Re: Test 2 #5

Calculate the difference in energy, deltaE.

Then apply E=hv and solve for v to get the frequency :)

PranithaPrasad
Posts: 58
Joined: Fri Sep 29, 2017 7:06 am
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### Re: Test 2 #5

But when I did that I got 2.92x10^15 for the energy and 4.41x10^48 for the frequency?

Christietan3F
Posts: 18
Joined: Sat Jul 22, 2017 3:01 am

### Re: Test 2 #5

Your delta E should be 1.937x10^-18, maybe that's what's messing up your answer?

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