Question

Calculate the wavelength transition in the Balmer series of atomic hydrogen.

Answer: -

1.52300 * (10^6) (m^(-1)) = 1 523 000 m^-1

Answer 1

@Zarkon @Yahoo! @FoolAroundMath @mathslover Please help:)

Answer 2

1.52300 * (10^6) (m^(-1)) = 1 523 000 m^-1

Answer 3

I want how to calculate it?

Answer 4

I just searched for net may be this helps u :
http://books.google.co.in/books?id=qQQcOhxHdxoC&pg=PA277&lpg=PA277&dq=Calculate+the+wavelength+transition+in+the+Balmer+series+of+atomic+hydrogen.&source=bl&ots=m2sdIauFHc&sig=3ymIqHXHQZe5JaTEbQmDVl1xF3s&hl=en&sa=X&ei=wEf4T46KNoXJrAeF7KjDBg&ved=0CEMQ6AEwAg

Answer 5

@maheshmeghwal9 see my calculation plzz

Answer 6

but where is that?

Answer 7

1.52300 * (10^6) (m^(-1)) = 1 523 000 m^-1

Answer 8

this is the answer of this question
I don't want this
I want solution of this question ending with this answer.

Answer 9

@Rohangrr Please see the question now:)

Answer 10

this is the actual question.

Answer 11

editing the question hmm...

Answer 12

ya

Answer 13

better go to this @maheshmeghwal9

Answer 14

http://en.wikipedia.org/wiki/Balmer_series

Answer 15

I have all the theories
but how to do this question?
the link doesn't help becoz it says same as my book.
no difference:/

Answer 16

The wavelengths for the Balmer series are all between 410 and 656 nm. So, I'm not sure what your answer is for, frequency? Okay, If you take 1/(your answer) you get the wavelength for the first Balmer transition from the Rydberg formula.

Answer 17

the same prob is with me that no
\[n_1 \space ; \space n_2\]& Z = atomic number are given
therefore i can't apply this formula
\[\frac{1}{\lambda}=R \times Z^2 \times [\frac{1}{n_1^2}-\frac{1}{n_2^2}].\]

Answer 18

n1 = 2, n2 = 3

Answer 19

how did u gt that?
wt is Z?

Answer 20

oops sorry Z=1
given in the question

Answer 21

but how did u gt this?
n1 = 2, n2 = 3

Answer 22

That's part of the definition of the Balmer series. All transitions end with the second orbital level: n = 2. The transition that has the wavelength for your answer starts from the third orbital level, so n2 = 3.

Answer 23

n2 may be greater than 3 too.

Answer 24

??????????

Answer 25

Yes. Look at the picture of the levels, under the Physics topic, here:
http://en.wikipedia.org/wiki/Hydrogen_spectral_series

Answer 26

Then why did u choose only n2=3
why not n2>3

Answer 27

Because if you flip your answer: 1/(your answer)
you get the wavelength from the transfer n2 = 3 to n1 = 2.

Answer 28

but if i didn't know my answer
then wht did u do?

Answer 29

Is the question giving you energy levels and asking for the wavelength?
I've had questions that ask for the wavelengths for all possible transitions. What is the question asking for?

Answer 30

The question is only this much.
Calculate the wavelength transition in the Balmer series of atomic hydrogen.



& one more thing i think if u flip my answer u will get wave number not wavelength becoz my answer is wavelength.

Answer 31

becoz \[\bar \nu = \frac{1}{\lambda}\]

Answer 32

I'm not sure what you mean by wave number. Your answer is the frequency for the first energy transition. So, if you flip it you get the wavelength.
There are wavelengths that result from a transition from n = 4 to n = 3, and from n = 5 to n = 2 ... there are many transitions and wavelengths possible.
If you just want one wavelength, I'd give the one we've been talking about, from n = 3 to n = 2.

Answer 33

ok i will see it
thanx for help:)

Answer 34

welcome