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/c

^{2}, so for an alpha particle (m =He4) we need 4u = 4 (931 MeV/c

^{2})

Then:

l

_{D}= (4.13 x 10

^{-`5}eV*s)/ 2[4 (931 MeV/c

^{2})] = 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 10

^{3 }mm) (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 = N

_{o}exp (-lt)

let t = T

_{1/2}the half life, then:

N = N

_{o}exp (-lT

_{1/2})

or N / N

_{o}= ½ = exp (-lT

_{1/2})

and: - ln 2 = (-lT

_{1/2})

T

_{1/2 }= (ln 2)/ l

5) The "

*mean life*" of one of the atoms in the sample will be:

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)

p = [2mT]

^{1/2}(see also #31)

A

*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

4m1

^{2}T1 (T1 - T2') = (m1 + m2)

^{2}T2' + 4m1

^{2}T1

^{2}- 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}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}

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 10

^{10}cm/sec) = 6.43 x 10

^{13}/s

11) Since the current decreases

*uniforml*y, we don't need di/dt and can instead make use of the finite difference: D(I)/ D(t)

And we know the voltage:

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 r

^{2}E1 = 4 π r(4 πr

^{2}/ 3)

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

^{2})

Therefore, E1 = qr/ a

^{3}

Further the associated energy is obtained via integration:

W1 = 1/ 8 π [∫ (0 to a) E1

^{2}(dA)

= 1/ 8 π [∫ (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 [∫ (a to oo) (q/ r

^{2})

^{2}(4 π r

^{2}) dr

W2 = q

^{2}/2 [∫ (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

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

__total impedance__is (bear in mind j = [-1]

^{-1/2}:

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})]

^{1/2}

w = 10

^{5}[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:

C

_{V}dT = C

_{p}- C

_{v }/ (¶V/ ¶T)

_{V}dV

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

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