Calculating Electric Potenial Difference

[chef99]

Homework Statement

In a TV tube, an electric potential difference accelerates electrons from a rest position towards a screen. Just before striking the screen, the electrons have a wavelength of 1.0 x10-11m. Find the electric potential difference.[/B]

Homework Equations

λ = h/mv

E_k = qΔV

The Attempt at a Solution

[/B]
First, determine the velocity

λ = h/mv

v = h/λm

v = 6.63 x10^-34J / (1.0 x10^-11m)(9.11 x10^-31m/s)

v = 7.28 x10^-7Now determine the electric potential difference

E_k = qΔV

1/2 mv² = eΔV

V = 1/2 mv² /eΔ

V = 1/2 (9.11 x10^-31kg)(7.28 x10^-7m/s)² / (1.60 x10^-19C)


V = 1.51 x10^-24

The electric potential difference is 1.51 x10^-24 V right before the electron strikes the screen.

I am fairly confident in my methods/answer but I haven't had to calculate the electric potential difference before so I just want to make sure I have the right idea. Any feedback would be great.

 

[Charles Link]

For the velocity, you put an extra minus sign in the exponent. It should be +7. Otherwise, your methodology is correct.

 

[chef99]

For the velocity, you put an extra minus sign in the exponent. It should be +7. Otherwise, your methodology is correct.

Oh wow, thank you for catching that.

So the electric potential difference is actually 1.51 x10^-4

 

[Charles Link]

Oh wow, thank you for catching that.

So the electric potential difference is actually 1.51 x10^-4

Try the arithmetic again. I get $V=1.51 \cdot 10^{+4}$ volts.