Question
Asked Oct 12, 2019

Recall Planck's constant equals 6.63 * 10-34 J*s and the speed of light is 3.00 * 10m/s.

Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 2 state.

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Expert Answer

Step 1

The values are given as follows:

Plank's consatant 6.63 x1034 J.s
speed of light 3.0x 10 m/s
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Plank's consatant 6.63 x1034 J.s speed of light 3.0x 10 m/s

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Step 2

The energy of the electron for the n quantum number can be expressed as follows:

Е, 3 -2.18х10-18 J
Here
n principle quantum number
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Е, 3 -2.18х10-18 J Here n principle quantum number

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Step 3

The difference between the electron’s energy from...

AE E - E
-2.18x1018 J
52
-2.18 x 10-18
J
22
1
=2.18x 1018 J
1
2252
21
J
100
2.18x 1018
=0.4578x 10-18 J
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AE E - E -2.18x1018 J 52 -2.18 x 10-18 J 22 1 =2.18x 1018 J 1 2252 21 J 100 2.18x 1018 =0.4578x 10-18 J

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