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user101
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Homework Statement
Find the kinetic energy of an electron in the lowest allowed energy state of a hydrogen atom.
Homework Equations
[tex]E = \frac{mv^2}{2} = \frac{mq^4}{2(4\pi\epsilon_0)^2n^2\hbar^2}[/tex]
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
[tex]E = \frac{mv^2}{2} = \frac{mq^4}{2(4\pi\epsilon_0)^2n^2\hbar^2}[/tex]
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 [tex]Epotential = Evacuum - \frac{q^2}{4\pi\epislon_0r}[/tex], but wasn't sure how to take into account the 10kV. Any help?