# Find frequency

1. Jul 30, 2011

### Saitama

1. The problem statement, all variables and given/known data
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)$v=\frac{2cRZ^2}{n^2}$

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

3. 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?

2. Jul 30, 2011

### 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.

3. Jul 30, 2011

### Saitama

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

4. Jul 30, 2011

### 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.

5. Jul 30, 2011

### Saitama

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:-
$$v=cRZ^2(\frac{201}{10000(10201)})$$

6. Jul 30, 2011

### pmsrw3

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.

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.

Last edited by a moderator: Apr 26, 2017
7. Jul 30, 2011

### Saitama

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?

Last edited by a moderator: Apr 26, 2017
8. Jul 30, 2011

### 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.

9. Jul 30, 2011

### Saitama

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

10. Jul 30, 2011

### pmsrw3

So use a calculator.

11. Jul 30, 2011

### Saitama

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

12. Jul 30, 2011

### 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.

13. Jul 30, 2011

### Saitama

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

14. Jul 30, 2011

### Saitama

But what's the correct method?

15. Jul 30, 2011

### I like Serena

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

16. Jul 30, 2011

### pmsrw3

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.

17. Jul 30, 2011

### Saitama

Thank you so much, i got it!!

18. Jul 30, 2011

### I like Serena

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

19. Jul 30, 2011

### Saitama

It becomes zero. :surprised

20. Jul 30, 2011

### I like Serena

Any idea why?
And how you should handle it?