Frequency of Electron Transition from (n+1) to n State

[Saitama]

Homework Statement

When an electron makes a transition from (n+1) state to nth state, the frequency of emitted radiations is related to n according to (n>>1):
(a)$v=\frac{2cRZ^2}{n^3}$
(b)$v=\frac{cRZ^2}{n^4}$
(c)$v=\frac{cRZ^2}{n^2}$
(a)$v=\frac{2cRZ^2}{n^2}$

Homework Equations

$$\frac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$$

The Attempt at a Solution

Solving the above equation and substituting the values i get:-
$$v=cRZ^2(\frac{1}{n^2}-\frac{1}{(n+1)^2})$$

$$v=cRZ^2(\frac{2n+1}{n^2(n+1)^2})$$

Now i am stuck, what should i do next?

 

[pmsrw3]

It's always a good idea when you're having trouble with a math problem to check whether you've used all the information given. You haven't used n>>1.

 

[Saitama]

It's always a good idea when you're having trouble with a math problem to check whether you've used all the information given. You haven't used n>>1.

I don't know how should i use n>>1?

 

[pmsrw3]

Well, try an example. Substitute in a big n, 100 for instance, and see what you get. Check whether it's closer to a, b, c, or d. Then try to figure out why.

 

[Saitama]

Well, try an example. Substitute in a big n, 100 for instance, and see what you get. Check whether it's closer to a, b, c, or d. Then try to figure out why.

That's a lot of calculation if i substitute n=100, and i am not able to solve it.
When i substitute n=100, i get:-
$$v=cRZ^2(\frac{201}{10000(10201)})$$

 

[pmsrw3]

if i substitute n=100, and i am not able to solve it.
When i substitute n=100, i get:-
$$v=cRZ^2(\frac{201}{10000(10201)})$$

There's nothing to solve! Just punch it into your calculator and see what you get. Or use http://www.google.com/landing/searchtips/#calculator". Then do the same for the four choices you're given.

That's a lot of calculation

No, that is not a lot of calculation. That's a small amount of easy calculations. You need to get used to calculating things if you're going to be taking science classes.

 

[Saitama]

There's nothing to solve! Just punch it into your calculator and see what you get. Or use http://www.google.com/landing/searchtips/#calculator". Then do the same for the four choices you're given.

Sorry, i can't use a calculator, this question is from my exam paper and in the examination room we were not allowed to use a calculator, so any other way to solve it?

 

[pmsrw3]

Are you in the exam room now? If so, you shouldn't be asking for help. If not, then you can use a calculator.

 

[Saitama]

Are you in the exam room now? If so, you shouldn't be asking for help. If not, then you can use a calculator.

No, don't talk silly, how can i be in the examination room and post a question?

 

[pmsrw3]

So use a calculator.

 

[Saitama]

Ok i did it using a calculator, i get my answer to be (a) option.

 

[pmsrw3]

Good. Now see if you can figure out why it comes out so close. If it's not obvious, try a few other n's.

 

[Saitama]

Hi Pranav-Arora!

If n >> 1, you can neglect small amounts in additions.
So for instance, n + 1 ≈ n
In the example where you calculated with n=100, you should e.g. replace 201 by 200.

Can you do that in your expression?

Hi!
If i solve it as you said, i again get (a) option.

 

[Saitama]

But what's the correct method?

 

[I like Serena]

My apologies pmsrw3, it was not my intention to interfere, so I deleted my post.

 

[pmsrw3]

But what's the correct method?

That's it. If you try a bunch of calculations, it should become obvious that when n is big, n is practically the same as n+1, and 2n is practically the same as 2n+1.

 

[Saitama]

That's it. If you try a bunch of calculations, it should become obvious that when n is big, n is practically the same as n+1, and 2n is practically the same as 2n+1.

Thank you so much, i got it!

 

[I like Serena]

Well, there is a catch...
What happens if you apply this estimation rule to your equation?

Solving the above equation and substituting the values i get:-
$$v=cRZ^2(\frac{1}{n^2}-\frac{1}{(n+1)^2})$$

 

[Saitama]

Well, there is a catch...
What happens if you apply this estimation rule to your equation?

It becomes zero.

 

[I like Serena]

Any idea why?
And how you should handle it?

 

[Saitama]

Any idea why?
And how you should handle it?

No idea.

 

[I like Serena]

Well, if you subtract 2 almost equal large quantities, the result is small but not zero.
This is something you can not neglect.
So (n+1) - n = 1 and this is not 0.

The trick is to eliminate subtractions, before you neglect stuff.

 

[Saitama]

Well, if you subtract 2 almost equal large quantities, the result is small but not zero.
This is something you can not neglect.
So (n+1) - n = 1 and this is not 0.

The trick is to eliminate subtractions, before you neglect stuff.

Thanks for clarification!
Can you please see to the thread "Atomic Structure Question" in "other Science".
tiny-tim replied that i should first find out the force, but i have never dealt with force in atomic structure.