Finding Kinetic Energy of an Electron in a Hydrogen Atom and a Cathode Ray Tube

[user101]

Homework Statement

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

Homework Equations

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

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:

Homework Statement

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.

Homework Equations

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

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?

 

[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?

 

[user101]

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?

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?