## Tuesday, January 18, 2011

This concludes the Advanced Physics test answers, for the questions given in:

31) The de Broglie wavelength is given by: L(D) = h/p

where h is the Planck constant and p is the momentum. (In this case, we need to use the version h = 4.13 x 10^-15 eV*s)

Since the problem information provides an energy E = 1 MeV, then we need p in the form:

p = [2mE]^½

These units mean it will also be easier to do the problem by using atomic mass units in MeV. Thus, 1 u = 931 MeV/c^2, so for an alpha particle (m =He4) we need 4u = 4 (931 MeV/c^2)

Then:

L(D) = (4.13 x 10^-15 eV*s)/ 2[4 (931 MeV/c^2)] = 10^-14 m

L(D) = 10^-12 cm

32) For a Young double slit, d sin (theta) = m L

gives the bright fringe, where m is the order and L the wavelength. Since the distance to the screen is much greater than the height of the bright fringe, sin (theta) = x / D, where x is the height and D is the distance to the screen. So:

delta (x) d/ D = L

delta(x) = DL/ d

then: delta(x) = (3 x 10^3mm) (6 x 10^-5 cm/ 0. 2 cm) = 0.9 mm

33) This application makes use of the energy-time uncertainty principle:

delta(E) delta(t) = h/ 2 π

Here, t = the mean lifetime of the excited state = T(½) = 8 x 10^-8 sec

And E= the half-width of the excited state = E(½)

then:

E(½) T(½) =h/ 2 π = 1.03 x 10^-34 J-s

E(½) = (1.03 x 10^-34 J-s)/ (8 x 10^-8 sec) = 1.2 x 10^-27 J

But: 1.6 x 10^-19 J = 1 eV (electron -volt)

so: E(½) = 10^-9 eV

34) We have: N = N(o) exp (-Lt)

let t = T(½) the half life, then:

N = N(o) exp (-L (T(½) )

or N/ N(o) = ½ = exp (-LT(½))

and: - ln 2 = - LT(½)

T(½) = (ln 2)/ L

35) The "mean life" of one of the atoms in the sample will be:

= INT (from 0 to oo) exp(-Lt) dt

= 1/ L {exp(-Lt) ] 0 to oo

= 1/L [exp (-L*0)] = 1/L [exp(o)] = 1/L

since exp(0) = 1

36) We need to apply both conservation of energy, and conservation of momentum.

For the first: T1 = T1' + T2'

for the second: P1 = P1'cos(theta) + P2' cos (phi)

and: 0 = P1'sin(theta) - P2' cos(phi)

A head-on collision implies phi = 0 and theta = 180 deg

so:

[2m1T1]^½ = ([2m1T1']^½ ) cos(180) + ([2m2T2']^½ ) cos (0)

But cos (0)= 1 and cos (180) = -1, so:

[2m1T1]^½ = - [2m1T1']^½ + [2m2T2']^½

Now, since T1 = T1' + T2':

2m1T1 + 4m1 [T1T1']^½ = 2m2T2' - 2m1T1'

4m1 [T1T1']^½ = - 2m1T1' + 2m2T2' - 2m1(T1 - T2')

whence:

2m1[T1(T1 - T2')]^½ = (m1 + m2)T2' - 2m1 T1

4m1^2 T1 (T1 - T2') = (m1 + m2)^2T2' + 4m1^2T1^2 - 4m1T1(m1 + m2)T2'

(m1 + m2)^2 T2'^2 - 4 m1m2T1T2' = 0

Then:

T2/T1 = 4 m1m2/ (m1 + m2)^2

37) We sketch the HCl molecule so:

H o<--------(x)---------- cm------------------->O (Cl)

And 1.27 Å defines the entire distance as shown (arrow to arrow) and let 'x' be the distance from the center of mass (cm) to the H atom. (Recall 1Å = 10^-8 cm).

Then:

x = 36(1.27Å - x)

37x = 36 (1.27 Å)

x = 36(1.27 x 10^-8 cm)/ 37

m(p) = 1.67 x 10^-24 g

The moment of inertia I is given by:

I = m(p)^2 x^2 + 36 m(p) [(1.27 x 10^-8 cm) - x]^2

I = (1.67 x 10^-24 g)(36/37)^2 + 36/(37)^2 [(1.27 x 10^-8 cm)]^2

I = 2.6 x 10^-40 g-cm^2

38) The Lande g-factor is defined:

g = [ J(J+ 1) + S(S +1) - L(L +1)/ 2J(J + 1) ] + 1

From our previous exposure to quantum mechanics

See, e.g.

http://brane-space.blogspot.com/2010/07/l-s-coupling-problem-solutions-and-more.html

and:

http://brane-space.blogspot.com/2010/07/more-quantum-problems.html

the 1D 3/2 state implies: L = 2, S = 1 and J = L + S = 2 + 1 = 3

then:

g = [3(3 + 1) + 1(1 + 1) - 2( 2+ 1)/ 2(3)(3 + 1)] + 1 = 1.33

39) Recall the nuclear relation:

f = M(A, Z) - A/ A

and that graphing this vs. A yields a constituent mass 1.0085A. Thus the total binding energy is: (0.0085A) x (931 MeV) or about 7.9 MeV per nucleon.

40) Here, it's important to recognize 2144 cm^-1 as the wave number, k. Also:

k = f/c = 2144 cm^-1

then the frequency:

f = kc = (2144 cm^-1) (3 x 10^10 cm/sec) = 6.43 x 10^13/s

41) Since the current decreases uniformly, di/dt -> delta(I)/ delta(t)

And V = L [delta(I)/ delta(t) ] = 0.25 H[ 2A/ (1/16s)]

V = 8 volts

42) Earth's atmosphere extends about 50 miles. If we assume the density of air to be constant:

p2 - p1 = rho g(z2 - z1)

where p2, p1 are the different pressures at heights z2 and z1, respectively, and rho is the air density with g the acceleration of gravity.

We can form a simple proportion based on the simplifying assumption:

(p2 - p1)/ p2 = (z2 - z1)/z2

p2 is what we need to find, and z2 is the height for 50 miles or ~ 50 x 5,000' = 250,000'.

whence:

(z2 - z1) = 200'

(p2 - p1)/ p2 = (z2 - z1)/z2 = 200'/ 250,000' = 0.0008

so:

(p2 - p1) = 0.0008 (p2)

p1 = p2 - 0.0008 (p2) ~ p2 ~ 76 Hg

43) For this solution, we let all quantities inside the sphere be denoted by (1) and all the quantities outside by (2).

By Gauss' law:

4 π r^2 E1 = 4 π rho(4 πr^2/ 3)

where rho (charge density) = 3 q / (4 π a^2)

Therefore, E1 = qr/ a^3

Further the associated energy is obtained via integration:

W1 = 1/ 8 pi [INT (0 to a) E1^2 (dA)

= 1/ 8 π [INT (0 to a) (qr/ a^3)^2 (4 π r^2) dr

W1= q^2/ 10a

By Gauss' law:

4 π r^2 E = 4 πq

E2 = q/ r^2

W2 = 1/ 8 π [INT (a to oo) (q/ r^2)^2 (4 π r^2) dr

W2 = q^2/2 [INT (a to oo) dr/ r^2]

W2 = q^2/ 2a

Therefore: the total electrostatic energy W = W1 + W2

W = q^2/ 10a + q^2/ 2a = 3q^2/ 5a

44) It helps here to sketch the circuit ( see graphic)

From this, the total impedance is (bear in mind j = [-1]^½):

Z_T = 1/ j wL + 1/ (1/ jw C1) + 1/ (1/ jwC2)

Z_T = 1/ jwL + jw( C1 + C2)

Z_T = j[ - 1/wL + w(C1 + C2)]

Z_T = -1/wL + w(C1 + C2) = 0 at the natural frequency

so:

w^2 = 1/ L(C1 + C2)

w = [1/ (10^-5) (30 x 10^-6)]^½

w = 10^5 [1/3]^½

w = 57,735 /s = 2 π f

so the natural frequency f= w/ 2 π= 9, 189 /s

or about 9.2 kilocycles per second

45) From thermodynamic relations we have:

C(v) dT = C(p) - C(v)/ (@V/ @T)_V dV

where the @ denote partials

Then:

(@T/@V)_s = (C(p)/ C(v) - 1) (@T/@V)_p

But:

(@T/@V)_p = T/V since T = PV/ nR

dT/ T + (C(p)/ C(v) - 1) (dV/V) = 0

Integrating both sides:

TV^(C(p)/ C(v) - 1) = k

PV^(C(p)/ C(v))/ nR = k

PV^(C(p)/ C(v)) = k/ nR

Then:

n = k/ PV^[C(p)/ C(v)] R