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Ok, I'm reviewing for my exam and I'm having issues with a few of the answers they have given... Let me know what you think. This practice exam seems a little iffy to me.... There seem to be mistakes in it all over. One question asked the angular momentum of the earth and said the answer was just 2/5MR^2 (no omega), which is just one half of the angular momentum, the rotational inertia.... I'm supposed to be helping a bunch of guys out preparing for this test too, so any help is greatly appreciated. Anyway...

First problem.

Initially a hydrogen atom is in its ground state . An electron with kinetic energy

10.6 eV collides with the atom and excites it. What is the remaining kinetic energy of the

electron?

A. 13.6 eV

B. 23.9 eV

C. 3.0 eV

D. 0.6 eV

E. 0.4 eV

-13.6/n^2 eV

Ground = -13.6 eV

n1 = -3.4 eV

n2 = -1.51 eV

So the electron would try to raise the energy state from the ground to the first energy state... correct (won't reach farther ones)? So -13.6/4 - (-13.6/1) = 10.2 eV, which leaves 0.4 eV left over for the electron as far as i can see, but the answer is supposed to be D - 0.6 eV

Never mind the second problem. We've been working with bike tires which are just MR^2, but I guess they wanted us to use a disk, 1/2MR^2, for this one. So I figured it out, there's just the first one now.

Thanks again for any help.

First problem.

## Homework Statement

Initially a hydrogen atom is in its ground state . An electron with kinetic energy

10.6 eV collides with the atom and excites it. What is the remaining kinetic energy of the

electron?

A. 13.6 eV

B. 23.9 eV

C. 3.0 eV

D. 0.6 eV

E. 0.4 eV

## Homework Equations

-13.6/n^2 eV

Ground = -13.6 eV

n1 = -3.4 eV

n2 = -1.51 eV

## The Attempt at a Solution

So the electron would try to raise the energy state from the ground to the first energy state... correct (won't reach farther ones)? So -13.6/4 - (-13.6/1) = 10.2 eV, which leaves 0.4 eV left over for the electron as far as i can see, but the answer is supposed to be D - 0.6 eV

Never mind the second problem. We've been working with bike tires which are just MR^2, but I guess they wanted us to use a disk, 1/2MR^2, for this one. So I figured it out, there's just the first one now.

Thanks again for any help.

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