Calculate vmax and voltage from X-ray distribution function

[VH1]

Homework Statement

The X-ray intensity distribution function for an X-ray lamp is
given on the figure. What is the maximum velocity of the electrons? What is the potential difference under which the lamp is operating? Figure: http://imgur.com/01kCBc8

Homework Equations

Kmax = 0.5m(vmax)^2
Kmax = hf - work function
λ = c/f
c = speed of light = 3 x 10^8 m/s
Peak wavelength (from the figure) = 3.5 x 10^-9 m
h = Planck's constant = 6.63 x 10^-34
m = mass of electron = 9.11 x 10^-31

The Attempt at a Solution

Doubt I'm starting at the right place.
0.5mv^2 = hf (work function??)
0.5mv^2 = h(c/λ)
v = sqrt [2h(c/λ)/m]
v = 1.1 x 10^7 m/s
Given answer is 1.5 x 10^7 m/s

As for the potential difference, I have no idea how to calculate it.

 

[gneill]

Hi VH1,

Welcome to Physics Forums!

X-Ray production isn't quite the same thing as the photoelectric effect. In this case electrons are stopped by a target and shed their energy in the form of photons.

Looking at the intensity distribution, which wavelength on the curve represents the highest energy photons (not the total intensity of emitted photons, but the highest photon energy)?

For the potential difference, how much energy does a charge gain by "falling through" a given potential difference? What's the formula?

 

[VH1]

Thanks gneill for the warm welcome and help - I managed to work out both questions with that.

Cheers