RI 20 Nuclear Physics tutorial solutions

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 Tutorial 20 Nu clear Physics Suggested Solutions DISCUSSION QUESTIONS Nuclear Reactions D1 (a) (i) α-particle comes to a stop at distance of closest approach and then reverses its motion along the same line (zero impact parameter b) (ii) α-particle experiences a large deflection when it passes close to the nucleus (small impact parameter b) (iii) α-particle experiences little deflection when it passes some distance from the nucleus (large impact parameter b) very small, very massive, positive nucleus (not to scale) RafflesInstitution (b) Observation Conclusion Most α-particles experienced little to no deflection Results as expected in Thomson’s plum - pudding model of atom. A small fraction of α -particles experienced large deflection ; a few α- particles are deflected at angles ° 170 . Large deflection possible only if the nucleus is massive (99.99% mass of atom). The very small fraction indicates that the nucleus is very small ( −− − 14 1510 10 m )

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 D2 (a) Neutrons are neutral and cannot be measured directly using electric or magnetic mass spectrometer. Note Experimentally, we can determine the energy or speed of a charged particle using either electric or magnetic mass spectrometer (for low energy) by measuring the amount of deflection, provided the mass of the particle is known. For high energy charged particles, an electromagnetic calorimeter is more appropriate. When a charged particle enters the calorimeter, it initiates a particle showers, i.e., the original charge particle transfers all its energy to other particles (via electromagnetic interaction). The energy of the particle shower is then deposited within the calorimeter and measured. https://www.desy.de/~garutti/LECTURES/ParticleDetectorSS12/L10_Calorimetry.pdf (b) Maximum recoil speed of the target nucleus occurs when the neutron collides head - on with the target nucleus. (c) Consider the head-on collision between a neutron and the target nucleus. mn m mn m vn v un before collision after collision Principle of conservation of momentum: = +nn nnmu mv m v  Since the collision is elastic, = −nnu vv ⇒ = −nnv vu  Substituting  into , ( )= −+nn n nm u m v u mv ⇒ += 2 n n n m v mvu m Since the target nuclei has mass 1m and 2m with recoil speed 1v and 2v , respectively, ++= =1 11 2 22 22 nn n nn mv mv mv mvu mm ⇒ −= − 11 22 21 n mv mvm vv

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 (d) The data for the collision are as follow: p 1.00 mu= ; 71 p 3.30 10 m sv −= × N 14.0 mu= ; 61 N 4.70 10 m sv −= × Substituting the date into the equation × ×− × ×= ×− × −×= =−× 76 67 7 7 1.00 3.30 10 14.00 4.70 10 4.70 10 3.30 10 3.28 10 1.16 2.83 10 n uum u u D3 Nuclear equation: →26 4 13 2H + Li 2 He Decrease in mass ∆ = + −× = 2.013553 6.013476 2 4.001505 0.024019 muu u u