bobsmith76
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Homework Statement
An isotope of hydrogen passes, without deflection, through a velocity selector that has an electric field of 2.40 × 10^5 N/C and a magnetic field of 0.400 T. It then enters a mass spectrometer that has an applied magnetic field of 0.494 T and consequently describes a circular path with a radius of 3.80 cm.
(a) What is the mass of the particle?
Homework Equations
v (velocity) = E/B
where E = electric field intensity
B = magnetic field intensity
.5mv^2 = qV
V = kq/r
The Attempt at a Solution
step 1. find the velocity
v = E/B
(2.4 * 10^5)/.4 = 6 * 10^5
step 2. find voltage
I tried to find q and V, then find the mass but I couldn't so I'm working backwards, starting with the answer and trying to find V
.5mv^2 = qV
.5* (5.01*10^-27)*(6*10^5)^2 = 1.6 * 10^-19 * V
step 3 get V by itself
(.5* (5.01*10^-27)*(6*10^5)^2)/(1.6 * 10^-19) = V
which is
1.44 * 10^20
step 4 look for an equation which will give me V (electric potential difference)
V = kq/r
((9*10^9)(1.6*10^-19)/.238 =
I forgot what that number is, but it did not add up to 1.44 * 10^20, so there's something I'm doing wrong. Let's try a different equation.
E (electric field intensity) = ΔV/Δd
step 5 E * d = V
The E with a magnetic field intensity of .494 is 2.9*10^5 so
2.9*10^5 * .238 =
I also forget what that number is but it also does not add up to around 10^20