Can someone help me solve this? In a Rutherford scattering experiment with iron

Question

Can someone help me solve this?

In a Rutherford scattering experiment with iron nuclei as the target the investigators measured the counting rate to be 3 events/s at 65 degree. Next the investigators replaced the iron target with tungsten. At what angle did they have to set the detector to get the same counting rate as they had seen with iron at 65 degree? Finally the investigators replaced the tungsten target with uranium. At what angle did they have to set the detector to get the same counting rate as they had seen with iron at 65 degree?

Explanation / Answer

The number of particles striking the detector per unit area is dependent on many factors like

thickness of the foil, Kinetic energy of the alpha particles, target to detector distance etc..

The formula which includes all these factors is

N() = NinLZ2k2e4 / 4r2KE2Sin4 (/2)

where N () = Number of particles per unit area striking the detector

Ni = Number of incident alpha particles

n= atoms per unit volume in target

L =thickness of target

Z= atomic number of target

e = electron charge

k = Coulomb's constant

r = target-to-detector distance

KE = Kinetic energy of alpha particle

= scattering angle

Now, obviously, so much data is not given in the question, which actually means that main of the conditions remain constant in all the three cases. For example, the KE of alpha particles, target-to-detector distance, and many such quantities remain unchanged because you are only changing the target.

Also, since the thickness' and atoms per unit volume are not given separately for the three targets, I am taking the freedom to assume them also to remain constant.

Now, if I replace all the constants and unaffected values with one single constant, say K, the equation simplifies to

N() = K Z2 / Sin4 (/2)

where K is some constant. So, as we see, what is causing the change is the atomic number Z.

If we want N () to be same in all the three cases, K Z2 / Sin4 (/2) also needs to be the same in all the cases.

So, K Z12 / Sin4 (1/2) = K Z22 / Sin4 (2/2) = K Z32 / Sin4 (3/2)

or  Z12 / Sin4 (1/2) = Z22 / Sin4 (2/2) = Z32 / Sin4 (3/2)

where Z1 = 26 ; Z2 = 74 ; Z3 = 92 ; 1 = 65 degrees

262 / Sin4 (65/2) = 742 / Sin4 (2/2) = 922 / Sin4 (3/2)

for Tungsten :

Sin4 (2/2) = 742 x Sin4 (65/2) / 262 = 0.6751229

Sin (2/2) = 0.906453875

2/2 = Sin-1 (0.906453875) = 65.01981298

2 = 130.039626 ~ 130 degrees

for Uranium :

Sin4 (3/2) = 922 x Sin4 (65/2) / 262 = 1.04350625

Sin (2/2) = 1.010703486

Note that Sin() can never be greater than 1, but here we are getting more than one. Does it mean that 2 is greater than 180 degrees? No, it actually means that if we use Uranium as the foil, we are never going to get the count as we got with the iron foil. Uranium being so highly charged reflects the alpha particles so powerfully that we cannot obtain the previous count.

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