A stream of protons, each with a speed of 0.8250c, are directed into a two-slit experiment where the slit separation is 2.00 10-9 m. A two-slit interference pattern is built up on the viewing screen. What is the angle between the center of the pattern and the second minimum (to either side of the center)?
p = mv
lambda = h/p
d*sin(theta) = (m + 1/2)*lambda
The Attempt at a Solution
I have tried to first solve for the wavelength in the experiment by using p = mv. With this I get:
p = (1.673E-27)*(0.8250)*(3E8)
p = 4.14E-19
Then I solve for the wavelength using lambda = h/p:
lambda = (6.63E-34) / (4.14E-19)
lambda = 1.6E-15
Once I have the wavelength, I use the double slit formula from Young's Experiment to try and calculate the angle, by using m = 1 and then solving for arcsin:
theta = arcsin ( m*lambda / d)
theta = arcsin ( 1.5*(1.6E-15) / (2E-9))
However this gives me a very small angle which obviously is the incorrect answer.
Am I approaching this completely wrong, or am I just goofing up somewhere?