## Friday, September 3, 2021

### Solutions To Physics Problems (Set 3)

1) 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]1/2

These units mean it will also be easier to do the problem by using atomic mass units in MeV. Thus, 1 u = 931 MeV/c2, so for an alpha particle (m =He4) we need 4u = 4 (931 MeV/c2)

Then:

l D = (4.13 x 10-`5 eV*s)/ 2[4 (931 MeV/c2)] = 10-14 m

l D = 10-12 cm

2) For a Young double slit, d sin (q) = 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 (q) = x / D, where x is the height and D is the distance to the screen. So:

D (x) d/ D = l

D(x) = Dl/ d

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

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

D(E) D(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

4) We have: N = No  exp (-lt)

let t = T1/2  the half life, then:

N = No  exp (-lT1/2)

or N / No   = ½ =  exp (-lT1/2)

and: - ln 2 = (-lT1/2)

T1/2   = (ln 2)/ l

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

(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

6) We need to apply both conservation of energy, and conservation of momentum. (There are two distinct angles involved, φ  and q)

For the conservation of energy: T1 = T1' + T2'

for the conservation of momentum : P1 = P1'cos(q) + P2' cos (φ)

and: 0 = P1'sin(q) - P2' cos(φ)

head-on collision implies the unique values:  φ = 0 and q = 180 deg

so:

[2m1T1]1/2 = ([2m1T1']1/2 ) cos(180) + ([2m2T2']1/2 ) cos (0)

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

[2m1T1]1/2 = - [2m1T1']1/2 + [2m2T2']1/2

Now, since T1 = T1' + T2':

2m1T1 + 4m1 [T1T1']1/2 = 2m2T2' - 2m1T1'

4m1 [T1T1']1/2 = - 2m1T1' + 2m2T2' - 2m1(T1 - T2')

whence:

2m1[T1(T1 - T2')]1/2 = (m1 + m2)T2' - 2m1 T1

4m12 T1 (T1 - T2') = (m1 + m2)2T2' + 4m12T12 - 4m1T1(m1 + m2)T2'

(m1 + m2)2 T2'2 - 4 m1m2T1T2' = 0

Then:

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

7) We sketch the HCl molecule, so:

H o<-------- ---------x--------->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 (mass of proton)

The moment of inertia I is given by:

I = m(p)2 x2 + 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-cm2

8) 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

9) 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.

10) 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 1010 cm/sec) = 6.43 x 1013/s

11) Since the current decreases uniformly, we don't need di/dt and can instead make use of the finite difference:  D(I)/  D(t)

And  we know the voltage:

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

Therefore:  V = 8 volts

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

p2 - p1 = r g(z2 - z1)

where p2, p1 are the different pressures at heights z2 and z1, respectively, and r 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

13) 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 r2 E1 = 4 π r(4 πr2/ 3)

where r (charge density) = 3 q / (4 π a2)

Therefore, E1 = qr/ a3

Further the associated energy is obtained via integration:

W1 = 1/ 8 π [ (0 to a) E12 (dA)

= 1/ 8 π [ (0 to a) (qr/ a3)2 (4 π r2) dr

W1= q2/ 10a

By Gauss' law:

4 π r2 E = 4 πq

E2 = q/ r2

W2 = 1/ 8 π [ (a to oo) (q/ r2)2 (4 π r2) dr

W2 = q2/2 [ (a to oo) dr/ r2]

W2 = q2/ 2a

Therefore: the total electrostatic energy W = W1 + W2

W = q2/ 10a + q2/ 2a = 3q2/ 5a

14) It helps here to sketch the circuit ( see graphic) and make use of angular frequency, w

From this, the total impedance is (bear in mind j = [-1]-1/2:

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

ZT = 1/ jwL + jw( C1 + C2)

ZT = j[ - 1/wL + w(C1 + C2)]

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

so:

w2 = 1/ L(C1 + C2)

w = [1/ (10-5) (30 x 10-6)]1/2

w = 105 [1/3]1/2

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

15) From thermodynamic relations we have:

V dT = C p - C / (V/ T) V dV

Then:

(T/V)s= (C / C v - 1) (T/Vp

But:

(¶TVp = T/V since T = PV/ nR

dT/ T + (C p/ C v - 1) (dV/V) = 0

Integrating both sides:

TV^ (C p/ C  - 1) = k

PV^ (C / C ) /  nR = k

PV^(C p/ C ) = k/ nR

Then:

n = k/ PV^[C p/ C ] R
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