# Atomic structure question

1. Jul 30, 2011

### Saitama

1. The problem statement, all variables and given/known data
A small particle of mass m moves in such a way that P.E.=-1/2mk(r)^2, where k is a constant and r is the distance of the particle from the origin. Assuming Bohr's model of quantization of angular momentum and circular orbit, r is directly proportional to:
(a)n2
(b)n
(c)$\sqrt{n}$
(d)none of these

2. Relevant equations
$P.E.=-\frac{KZe^2}{r}$

3. The attempt at a solution
I don't understand how should i start?

Last edited: Jul 30, 2011
2. Jul 30, 2011

### tiny-tim

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

### Saitama

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

### I like Serena

The force is the derivative of the potential energy.
That is, the derivative of the P.E. given in the problem statement, not the one you listed as relevant equation.

5. Jul 30, 2011

### Saitama

Why i have to find the derivative of P.E. given in the problem statement, not the one in the relevant equations?
And with what respect do i have to find the derivative?

Did you get that i refer P.E. as "Potential Energy"?

6. Jul 30, 2011

### I like Serena

The P.E. in the relevant equations is for charges attracting each other, which is what Bohr used in his model of the atom (btw, there should be a minus sign included).

The P.E. in the problem statement is about a different system, for instance a system with a mass on a spring, or a charge inside a sphere with homogeneous charge density.

Edit: The derivative is with respect to the only variable you have, which is r.

7. Jul 30, 2011

### Saitama

But how do i find the derivative?
And how do i calculate the force?

8. Jul 30, 2011

### I like Serena

You do know what a derivative is?

Start by listing the P.E. and the derivative of the P.E. with respect to r?
The latter is the force.

9. Jul 30, 2011

### Saitama

Yes, i know what a derivative is.

$$\frac{d}{dr}(P.E.)=-mkr$$

Am i right..?

10. Jul 30, 2011

### I like Serena

Yes! :)

11. Jul 30, 2011

### Saitama

But what next?

12. Jul 30, 2011

### I like Serena

What did tiny-tim suggest?

13. Jul 30, 2011

### Saitama

But i have never studied centripetal acceleration....

14. Jul 30, 2011

### I like Serena

All right, but then you will also get into trouble with angular momentum and with quantization....

What do you know about the mechanics of circular motion?

15. Jul 30, 2011

### Saitama

Nothing.
I haven't still reached to the circular motion.

16. Jul 30, 2011

### Saitama

I read circular motion on Wikipedia, i have learnt about uniform circular motion.

17. Jul 30, 2011

### I like Serena

So can you follow up on tiny-tim's suggestion?

"use centripetal acceleration to find the relation between v and r"

18. Jul 30, 2011

### Saitama

Is the knowledge of uniform circular motion sufficient for centripetal acceleration?

19. Jul 30, 2011

### Saitama

I think i have got the answer now.
I did it like this:-
$$\frac{d}{dr}(P.E.)=-mkr$$
$$ma_c=-mkr$$
$$-\frac{mv^2}{r}=-mkr$$
$$v^2=kr^2$$

Using Bohr's model of quantization of angular momentum,
$$mvr=\frac{nh}{2\pi}$$
$$v=\frac{nh}{2\pi mr}$$

Substituting this value of v in our previous equation, i get:-
$$\frac{n^2h^2}{4\pi^2m^2r^2}=kr^2$$
$$\frac{n^2h^2}{4\pi^2m^2}=kr^4$$

Therefore,
$$r^4\propto n^2$$
$$r^2\propto n$$
$$r\propto\sqrt{n}$$

20. Jul 31, 2011

*bump*