Modern Physics- Problem Set 2

Modern Physics for Engineers

Problem Set 2

Problem Set 2 - Due in class, Wednesday, Oct. 6, 1999.

SM&M 1-34.

Solution:

	KE = 5 mc^2
	E = KE + mc^2 = 6 mc^2
	gamma = E/mc^2 = 6   beta = Sqrt[1 - 1/gamma^2] = .972
	v = .972 c

SM&M 1-39. Before answering part (b), calculate the amount of energy release in the reaction:
4 H ---> He + 2 Nu_e
where H and He are hydrogen and helium atoms respectively and Nu_e is a massless particle. This is the primary energy producing reaction in the sun. Note: the masses given in the back of the book are the masses for the atoms. Thus the mass for H is the mass of a hydrogen atom and not just a proton.
Solution - Part A:
```
	            4x10^26 joules/sec 
	m = E/c^2 = ------------------ = 4.44 x 10^9 kg/sec
	            9x10^16 m^2/sec^2
```
Solution - question above:
```
	4 m_H c^2 - m_He c^2
	-------------------- = .008
	     4 m_H c^2
```
or .8% of the mass is available for the production of thermal energy.
Solution - Part B:
```
	.008 x 2x10^30 kg
	----------------- = 3.6x10^18 sec = 113 billion years
	4.44x10^9 kg/sec
```
Actually the sun will explode as a supernova in a fraction of that time.

SM&M 1-42. Also calculate how much kinetic energy each particle receives.

Problem Solution:

	KE = m_Ra c^2 - m_Rn c^2 - m_He c^2
	   = (226.0254 - 222.0175 - 4.0026) u 1.66x10^-27 kg/u 9x10^16 m^2/sec^2
	   = 7.9x10^-13 joules = 4.9 MeV

Distribution of kinetic energy:

	m_0 = m_Ra
	m_1 = m_Rn
	m_2 = m_He

	m_0 c^2 = E_1 + E_2 
	        = Sqrt[(p_1 c)^2 +(m_1 c^2)^2] + Sqrt[(p_2 c)^2 +(m_2 c^2)^2]

	0 = p_1 + p_2 --> p_1 = - p_2 = p

-->	(m_0 c^2 - Sqrt[(p c)^2 +(m_1 c^2)^2])^2 = (p c)^2 + (m_2 c^2)^2
	(m_0 c^2)^2 -2 m_0 c^2 Sqrt[(p c)^2 +(m_1 c^2)^2] = (m_2 c^2)^2
	                                   (m_0 c^2)^2 + (m_1 c^2)^2 - (m_2 c^2)^2
--->	E_1 = Sqrt[(p c)^2 +(m_1 c^2)^2] = ---------------------------------------
	                                               2 m_0 c^2
	                       (m_0 c^2 - m_1 c^2)^2 - (m_2 c^2)^2
	KE_1 = E_1 - m_1 c^2 = ----------------------------------- = .086 MeV
	                                  2 m_0 c^2
	                                               2 m_0 c^2
	                       (m_0 c^2 - m_2 c^2)^2 - (m_1 c^2)^2
	KE_2 = E_2 - m_2 c^2 = ----------------------------------- = 4.814 MeV
	                                  2 m_0 c^2
	KE_tot = KE_1 + KE_2 = m_0 c^2 - m_1 c^2 - m_2 c^2

The He carries away 98% of the kinetic energy.

SM&M 1-46.

Energy/momentum relationships

	E_e = Sqrt[(p_e c)^2 + (m_e c^2)^2]
	E_g1 = p_g1 c
	E_g2 = p_g2 c

Energy conservation:

	E_i = E_f
	E_e + m_p c^2 = E_g1+E_g2
	m_e = m_p = .511 MeV/c^2
	E_e = 1 MeV + .511 MeV = 1.511 MeV
	E_i = 2.022 MeV

Y momentum conservation:

	0 = p_g1 Sin[theta] + p_g2 Sin[-theta]
-->	p_g1 = p_g2 = p

X momentum conservation:

	px_i c = px_f c
	px_i c = Sqrt[E_e^2 - (m_e c^2)^2] = 1.422 MeV
	px_f = p_g1 Cos[theta] + p_g2 Cos[-theta]
	     = 2 p Cos[theta]

Solution:

	E_i = 2 p c --> pc = E_g = 1.011 MeV
	Cos[theta] =px_i/2 p = 1.422/2.022 = .703
-->	theta = 45.3 degrees

SM&M 1-47. Solution, Part A:

	p = r q B
	pc = r q B c
	   = .344 m  x 1.6x10^-19 coul x 2 T x 3x10^8 m/sec
	   = 3.3x10^-11 joules = 206 MeV
	E = Sqrt[(pc)^2 + (m c^2)^2]
	  = Sqrt[206^2 + 140^2] = 249 MeV
	       pc
	beta = -- = .827 ---> v = .827 c
	        E

Solution, Part B:

	E_i = E_f
	m_K c^2 = 2 E = 2 x 249 MeV = 498 MeV
	m_K = 498 MeV/c^2