3. (6 points) 3.1 What would be the approximate counting...

3. (6 points) 3.1 What would be the approximate counting rate observed in the Rutherford scattering of 10 MeV \alpha-particle off lead foil at an angle of $\theta = \frac{\pi}{2}$ in the laboratory? • Incident flux: $10^6$ \alpha-particles per second on the foil • Lead foil: Thick: 0.1 cm, Density: 11.3 g/cm³ • Detector: Distance from interaction point: 100 cm, Transverse area 1 cm x 1cm 3.2 What would be the counting rate at $\theta = 5^\circ$? 3.3 By about how much would your answer change if the above angles were specified for the center of mass be quantitative but use approximation where necessary.

Transcript

00:01 Hello students, in this question we have given alpha particle of kinetic energy ke equals to 6 mega electron volt and it is inciting at the rate, suppose r -dash it is equals to 3 .0 mid -pladbit 10 to the power 7 per second and on the gold foil having z equals to 79 on a thickness t equals to 3 .0 mid -pladbit 10 to the power minus 6 meter.

00:27 Okay, and a circular detector of diameter small d is equal to 1 .0 centimeter is placed at a distance small r equals to 12 centimeter at an angle theta equals to 30 degree with the direction of incident alpha particles.

00:42 So we have to determine the rate does the detector measure the scattered alpha particles.

00:48 Okay.

00:48 So we can write that number of electron density small n, it is equals to avogato number matter, divided by density divided by molecular weight.

00:57 So i've got row number is 6 .02 mudplied weight 10 to the power 23 molecules per mole, mudplared by density of the gold which is 19 .3 gram per mole divided by molecular weight which is 197 gram per mole.

01:11 So from here we get 5 .9 mudplad weight 10 to the power 28 atoms per meter cube.

01:20 Now we can calculate from the rutherford scattering formula the probability per unit area so n theta this is given by small n mudplated by t divided by 4 r square but plight by small z capital z divided by two kinetic energy and this whole square smaller z is the atomic number for the helium which is 2 okay and mudplied by e squared divided by 4 pi epsilon not and this whole square and modipleared by 1 by sine to the power 4, theta by 2.

01:54 So substituting all the values, so we get probability n theta, it is equals to small n which is this value.

02:00 So 5 .9, mudplared bit and to the power 28, divide by 4, muddplared by radius, which is from here 1 by 2 cm, so 1 by 2 muddbletabbit 10 to the power minus 2 meter and this whole is square.

02:13 Multiplied by small z which is 2 capital z which is 79 divided by 2 but played by kinetic energy which is 6 mega electron volt and this whole square but plight by e square by 4 pi a epsilon not it has value 1 .44 mega electron volt so whole square by by by 1 by sign to the power 4 and theta which is 30 degree okay so sign 30 degree divided by 2 okay so from here after solving, we will get probability n theta, this is equal to 0 .246 per meter square.

02:49 Now we can calculate the total probability n theta multiplied by area...

Question

Please give Ace some feedback