# Kinetic Energies

1. Sep 2, 2007

### user101

1. The problem statement, all variables and given/known data

Find the kinetic energy of an electron in the lowest allowed energy state of a hydrogen atom.

2. Relevant equations

$$E = \frac{mv^2}{2} = \frac{mq^4}{2(4\pi\epsilon_0)^2n^2\hbar^2}$$

3. The attempt at a solution

m = 9.11* 10^(-31) kg
q = 1.6 * 10^(-19) C
pi = 3.14
n = 1
hbar = 6.59 * 10^(-16) eV * s

Are the values I chose correct?

Next Problem:

1. The problem statement, all variables and given/known data

Find the kinetic energy of a free electron, initially at rest at the back of a cathode ray tube, accelerated through a potential of 10kV to strike the phosphor layer.

2. Relevant equations

$$E = \frac{mv^2}{2} = \frac{mq^4}{2(4\pi\epsilon_0)^2n^2\hbar^2}$$

3. The attempt at a solution

I'm not too sure how to relate KE and potential.

I know that Total Energy = Potential E + Kinetic E, but I don't know Total Energy in order to use that generalized equation.

The next thing I thought was to use $$Epotential = Evacuum - \frac{q^2}{4\pi\epislon_0r}$$, but wasn't sure how to take into account the 10kV. Any help?

2. Sep 2, 2007

### EricVT

For the second question, the kinetic energy gained by an electron passing through 1 volt of electrostatic potential is 1 eV, so a 10kV potential would yield an increase in kinetic energy of 10keV, right? What is the energy of the free electron before being accelerated through the potential?

3. Sep 2, 2007

### user101

Well, for the electron BEFORE being accelerated would have no KE value. So, this would mean that the only KE gain would be the 10keV you described above?