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Finally! Anyone?
okay so the eqn is
(hc/lambda) = B + eV
If V is one the y-axis and 1/lambda on the x-axis
write it into the form y = mx + c
where y is V and x = 1/lambda
So:
eV = (hc/lambda) - B
Divide both sides by e to make V the subject
V = [ ( hc/lambda) - B ] /e
so V = [ (hc/e) x 1/lambda ] - B/e
therefore gradient is hc/e
y - intercept is -B/e