Physics, 09.11.2019 03:31 taliajohnsom9901

Singly charged uranium-238 ions are accelerated through a potential difference of 2.00 kv and enter a uniform magnetic field of magnitude 1.20 t directed perpendicular to their velocities.
(a) determine the radius of their circular path.
(b) repeat this calculation for uranium- 235 ions.
(c) what if? how does the ratio of these path radii depend on the accelerating voltage?
(d) on the magnitude of the magnetic field?

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Answer from: gototiger

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Explanation:

First, we define the variables we are going to use

let:

$\Delta V = 2000 [V]$ the voltage difference

$q = 1.6*10^{-19} [C]$ the charge

$m = 238*1.6*10^{-27} [Kg]$ be the mass of U238, it comes from multiplying the isotopes mass in amu's by the mass of a proton

B = 1.20 [T]

We start by calculating the velocity of the ions, work is equal to charge times potential difference

$W = q\Delta V$ , which is nothing else than energy (the units are the same, Joules)

Therefore the kinetic energy of the particle will be equal to the magnitude of $q\Delta V$

let's find the velocity now :

$\frac{1m*v^2}{2}=q\Delta V = v = \sqrt{\frac{2q \Delta V}{m} }$

we get v =40996 [m/s]

a) We note that the magnetic field is perpendicular to the velocity of the ions, therefore we can use the lorentz force formula

$F = qvBsin( \theta)$ , where theta is the angle between B and V (which is 90 as they are perpendicular to each other), F is the centripetal force $\frac{mv^2}{r}$

so we get $qvBsin( 90) = \frac{mv^2}{r} = qB = \frac{mv}{r}$

we just replace v! => $r = \frac{mv}{qB} = r = \frac{m*(\sqrt{\frac{2q \Delta V}{m} })}{qB}$

m enters the square root and cancels out the inner m, and so does q.

we get now an expression for the radius of the circular path:

$r = \frac{1}{B} *(\sqrt{\frac{2m \Delta V}{q} })$

replacing values we obtain r = $\frac{1}{1.6}* \sqrt{\frac{2*238*1.6*10^-{27}*2000}{1.6*10^-19} } = 0.0609 [m]$

b) for uranium 235 we just replace the 238 with 235:

r = $\frac{1}{1.6}* \sqrt{\frac{2*235*1.6*10^-{27}*2000}{1.6*10^-19} } = 0.0605 [m]$

c) it depends heavily on the accelerating voltage, we just saw that the radius of the path does not change a lot with a change in the mass (0.0605 vs. 0.0609), if we drop the voltage difference to 1.00Kv we get for the 238 uranium ions a radius of 0.043, for the 235 uranium 0.042 which is approximately equal, the main factor is the voltage difference

d) it also depends heavily on the magnitude of the magnetic field, if we increase 1.6 T to let's say 2T we get a radiius of 0.034 for U-238. The decissive factor here is the angle theta between V and B, we got lucky here because they are perpendicular, in any other case smaller angles will decrease the magnitude of B.

Answer from: Quest

The answer would be a) -80of

Answer from: Quest

That depends on what kind of environ meant it is in in space it has the same force that was applied when the velocity was gained

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Singly charged uranium-238 ions are accelerated through a potential difference of 2.00 kv and enter...

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