- #1

- 565

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E = 54.4/n^2 eV

Use this to calculate the wavelength of the first line of the balmer series for singly ionized helium

E=54.4 for the first line. Whats the next step?

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- Thread starter Ry122
- Start date

- #1

- 565

- 2

E = 54.4/n^2 eV

Use this to calculate the wavelength of the first line of the balmer series for singly ionized helium

E=54.4 for the first line. Whats the next step?

- #2

Dick

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- #3

- 565

- 2

can you tell me what equation i need to use?

- #4

Dick

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You just gave it. Balmer lines are transitions between n>=3 and n=2. E=54.4eV isn't a line. It's the first energy level. Lines are the difference between energy levels.

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- #5

- 565

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- #6

Dick

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What's the relation between energy of a photon and it's wavelength?

- #7

- 565

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Balmer lines are transitions between n>=3 and n=2

So what value do I use for n, 2?

- #8

Dick

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- #9

- 565

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You don't use a value for n

if n isn't assigned a value, how can i find E?

if n isn't assigned a value, how can i find E?

- #10

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You need to calculate the energy levels for two values of n.

The difference between these energy levels is what you are interested in.

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