It's now time to provide the answers to the advanced physics question given in the earlier blog:

http://brane-space.blogspot.com/2011/01/advanced-physics-questions.html

1) We make use of the "photo-electric effect" here and note that the photo-electric threshold means that electromagnetic energy of less energy (that is, longer wavelength) will not possess sufficient energy to dislodge an electron from a material.

For the photo-electric effect: E = hf - W

Where W is the "work function" and W = hc/ L(t), and f is the frequency of the light.

L(t) denotes the photo-electric threshold, c is the speed of light and h is the Planck constant.

We solve via the standard energy equation:

E = hc [ 1/L - 1/L(t)]

noting the frequency f ~ 1/L (e.g. inversely proportional to the wavelength, L)

It helps considerably at this stage if one knows the quickie conversion factor (bearing in mind the GRE is a timed test) such that:

hc = 12.4 x 10^3 eV*A

then,

E = 12.4 x 10^3 eV*A [ 1/1800A - 1/ 2300A] = 1.498 eV

or, ZE = 1.5 eV

2) From the Bohr model of the atom, the applicable energy here would be given by:

E(n) = - (2 pi^2 me^4/h^2)(Z^2/n^2)

and since we know n = 1 and Z = 11, with m the mass of the electron, this reduces to:

E(n) = - 121(2 pi^2 me^4/h^2)

3) Using Cartesian coordinates:

r^ = x(i^) + y(j^) = z(k^)

and the Divergence (DIV) is defined on this basis:

Div (r^) = i^ (@/@x) + j^(@/@y) + k^(@/@z)

where @ denotes a partial differential

Then, DIV (r^) = @(x)/@x + @(y)/@y + @(z)/@z = 1 + 1 + 1 = 3

4) Here, we have:

and - as before: r^ = x(i^) + y(j^) = z(k^)

Then:

Then the gradient, grad (

i^ @(a_x(x))/@x + j^2 @(a_y(y))/@y + k^ @(a_z(z))@z

Then, grad (

5) a) Consider the matrix:

(0........0........1)

(0........1.........0)

(1.........0 .......0)

Then the Trace is the sum of the diagonal elements from the top left to the bottom right, or:

Tr = 0 + 1 + 0 = 1

5 b)

As we previously saw (while working on linear systems of homogeneous DEs), the eigenvalues wil be such that (assuming M is the matrix above):

Det[(M) - LI] = 0 where I =

(1....0.....0)

(0....1.....0)

(0.....0.....1)

Then we can write: 0 =

(-L .........0.........1)

(0.........(1 - L)....0)

(1 ...........0...........-L)

For which we obtain a characteristic eqn.

L^2(1 - L) - (1 - L) = 0, or

(1 - L) (L^2 - 1) = 0

yielding eigenvalues: L = 1, L = +/- 1

6) We have: F = N(U1 + 2i(U2))

where U1, U2 are orthonormal functions.

Then: F* = N(U1* - 2i(U2*))

So, FF* = [N(U1 + 2i(U2))][N(U1* - 2i(U2*))] = 1

(Refer back to our earlier work - last year- with complex arithmetic and complex conjugates!)

Whence:

U1U1* = 1 and U2U2* = 1

But:

U1U2* = U2 U1* = 0

Thus:

N^2 (U1 + 2i U2)(U1* - 2i (U2*) = 1

Therefore:

5N^2 = 1

So: N = (5)^-½

7) See sketch given for Problem (7)

From the set up (sketch) and the principle of moments:

F(L/2) - (W/2)(L/4) + (W/2)(L/2)

Bear in mind that prior to the release both men are supporting the plank equally, so the one we're interested in feels a force W/2 initially. Immediately after release, however, one can regard the plank as undergoing a rotation about its center of mass (which is exactly at its geometric center, since it's uniform)

Then: F = 3W/4 (after)

So the change in force experienced is:

¼ (3W - 2W) = ¼ W

8) This is: di^/dt = w x i^

9) d^2 (i^)/dt^2 = d/dt(di^/dt) = d/dt(w x i^) = w x w x i^

10) The reader ought to first note easily that the "density" is actually the mass density for the strings in units of grams per centimeter! (Not g/ cc!)

He also needs to realize that the velocity v in the two strings will be:

v = (τ /φ) where φ is the mass (per length) density and τ is the tension.

The "index of refraction" will then be: n = v2/ v1

So, by analogy with the reflection coefficient of optics:

R = (n - 1)^2/ (n + 1)^2

Substituting for n (using velocities v1, v2):

R = [(v2/v1) - 1]^2 / [(v2/ v1) + 1]^2

R = (v2 - v1)^2/ (v2 + v1)^2

Then we may compute R from mass densities alone:

R = {(1/ (φ2)^1/2 - 1/(φ1)^1/2)/ {(1/ (φ2)^1/2 + 1/(φ1)^1/2)}

Simplifying:

R = [(φ1)^1/2 - (φ2)^1/2/ φ1)^1/2 + (φ2)^1/2]^2

R = (5 g/cm - 3 g/cm)^2/ (5 g/cm + 3 g/cm)^2 = (2 g/cm)^2/ (8 g/cm)^2

Therefore, R = 4/ 64 = 1/16

11) The solution for this is shown in the second graphic inset above.

12) We use the relativistic form:

E= mc^2[1 - (v/c)^2]^-½ - 1]

Noting the quickie conversion factor: mc^2 = 0.511 MeV

Now, if E = 0.25 MeV, then:

0.25 MeV = 0.511 MeV [1 - (v/c)^2]^-½ - 1]

and:

3/2 = [1 - (v/c)^2]^-½

square both sides:

9/4 = 1/[1 - (v/c)^2]

Or: [1 - (v/c)^2] = 4/9

So: (v/c)^2 = 1 - 4/9 = 5/9

whence: (v/c) = (5/9)^½ = 0.75

so, v = 0.75c

13) We use: E = mc^2

and: L = hc/E for a massless particle

Then: mc^2 = hc/L

and m = h/cL

L = 6000A = 600 nm = 6 x 10^-7 m

h = 6.62 * 10^-34 J*s

c = 3 x 10^8 m/s

So:

m =3.7 x 10^-33 g

14) We have the power (1300 W) related to the E-field intensity via:

P = E^2(A)/ (2 u(o) c)

where A = 1 m^2 (area) and u(o) is the magnetic permeability = 4 pi x 10^-7 H/m

Then:

E = [P*(2 u(o) c) ]^1/2 = [1300 J/s*4 pi x 10^-7 H/m * 3 x 10^8 m/s]^1/2

= 700 V/m

15) From the Poynting vector: S = (EB)/u(o)

so: B = E/c

= (700 V/m)/ (3 x 10^8 m/s) = 2.3 x 10^-6 Wb/m^2 (weber per meter squared)

http://brane-space.blogspot.com/2011/01/advanced-physics-questions.html

1) We make use of the "photo-electric effect" here and note that the photo-electric threshold means that electromagnetic energy of less energy (that is, longer wavelength) will not possess sufficient energy to dislodge an electron from a material.

For the photo-electric effect: E = hf - W

Where W is the "work function" and W = hc/ L(t), and f is the frequency of the light.

L(t) denotes the photo-electric threshold, c is the speed of light and h is the Planck constant.

We solve via the standard energy equation:

E = hc [ 1/L - 1/L(t)]

noting the frequency f ~ 1/L (e.g. inversely proportional to the wavelength, L)

It helps considerably at this stage if one knows the quickie conversion factor (bearing in mind the GRE is a timed test) such that:

hc = 12.4 x 10^3 eV*A

then,

E = 12.4 x 10^3 eV*A [ 1/1800A - 1/ 2300A] = 1.498 eV

or, ZE = 1.5 eV

2) From the Bohr model of the atom, the applicable energy here would be given by:

E(n) = - (2 pi^2 me^4/h^2)(Z^2/n^2)

and since we know n = 1 and Z = 11, with m the mass of the electron, this reduces to:

E(n) = - 121(2 pi^2 me^4/h^2)

3) Using Cartesian coordinates:

r^ = x(i^) + y(j^) = z(k^)

and the Divergence (DIV) is defined on this basis:

Div (r^) = i^ (@/@x) + j^(@/@y) + k^(@/@z)

where @ denotes a partial differential

Then, DIV (r^) = @(x)/@x + @(y)/@y + @(z)/@z = 1 + 1 + 1 = 3

4) Here, we have:

**= a_x(i^) + a_y(j^) + a_z(k^)***a*and - as before: r^ = x(i^) + y(j^) = z(k^)

Then:

***r = a_x(x) + a_y(y) + a_z(z)***a*Then the gradient, grad (

***r) =***a*i^ @(a_x(x))/@x + j^2 @(a_y(y))/@y + k^ @(a_z(z))@z

Then, grad (

***r) =***a**a*5) a) Consider the matrix:

(0........0........1)

(0........1.........0)

(1.........0 .......0)

Then the Trace is the sum of the diagonal elements from the top left to the bottom right, or:

Tr = 0 + 1 + 0 = 1

5 b)

As we previously saw (while working on linear systems of homogeneous DEs), the eigenvalues wil be such that (assuming M is the matrix above):

Det[(M) - LI] = 0 where I =

(1....0.....0)

(0....1.....0)

(0.....0.....1)

Then we can write: 0 =

(-L .........0.........1)

(0.........(1 - L)....0)

(1 ...........0...........-L)

For which we obtain a characteristic eqn.

L^2(1 - L) - (1 - L) = 0, or

(1 - L) (L^2 - 1) = 0

yielding eigenvalues: L = 1, L = +/- 1

6) We have: F = N(U1 + 2i(U2))

where U1, U2 are orthonormal functions.

Then: F* = N(U1* - 2i(U2*))

So, FF* = [N(U1 + 2i(U2))][N(U1* - 2i(U2*))] = 1

(Refer back to our earlier work - last year- with complex arithmetic and complex conjugates!)

Whence:

U1U1* = 1 and U2U2* = 1

But:

U1U2* = U2 U1* = 0

Thus:

N^2 (U1 + 2i U2)(U1* - 2i (U2*) = 1

Therefore:

5N^2 = 1

So: N = (5)^-½

7) See sketch given for Problem (7)

From the set up (sketch) and the principle of moments:

F(L/2) - (W/2)(L/4) + (W/2)(L/2)

Bear in mind that prior to the release both men are supporting the plank equally, so the one we're interested in feels a force W/2 initially. Immediately after release, however, one can regard the plank as undergoing a rotation about its center of mass (which is exactly at its geometric center, since it's uniform)

Then: F = 3W/4 (after)

So the change in force experienced is:

¼ (3W - 2W) = ¼ W

8) This is: di^/dt = w x i^

9) d^2 (i^)/dt^2 = d/dt(di^/dt) = d/dt(w x i^) = w x w x i^

10) The reader ought to first note easily that the "density" is actually the mass density for the strings in units of grams per centimeter! (Not g/ cc!)

He also needs to realize that the velocity v in the two strings will be:

v = (τ /φ) where φ is the mass (per length) density and τ is the tension.

The "index of refraction" will then be: n = v2/ v1

So, by analogy with the reflection coefficient of optics:

R = (n - 1)^2/ (n + 1)^2

Substituting for n (using velocities v1, v2):

R = [(v2/v1) - 1]^2 / [(v2/ v1) + 1]^2

R = (v2 - v1)^2/ (v2 + v1)^2

Then we may compute R from mass densities alone:

R = {(1/ (φ2)^1/2 - 1/(φ1)^1/2)/ {(1/ (φ2)^1/2 + 1/(φ1)^1/2)}

Simplifying:

R = [(φ1)^1/2 - (φ2)^1/2/ φ1)^1/2 + (φ2)^1/2]^2

R = (5 g/cm - 3 g/cm)^2/ (5 g/cm + 3 g/cm)^2 = (2 g/cm)^2/ (8 g/cm)^2

Therefore, R = 4/ 64 = 1/16

11) The solution for this is shown in the second graphic inset above.

12) We use the relativistic form:

E= mc^2[1 - (v/c)^2]^-½ - 1]

Noting the quickie conversion factor: mc^2 = 0.511 MeV

Now, if E = 0.25 MeV, then:

0.25 MeV = 0.511 MeV [1 - (v/c)^2]^-½ - 1]

and:

3/2 = [1 - (v/c)^2]^-½

square both sides:

9/4 = 1/[1 - (v/c)^2]

Or: [1 - (v/c)^2] = 4/9

So: (v/c)^2 = 1 - 4/9 = 5/9

whence: (v/c) = (5/9)^½ = 0.75

so, v = 0.75c

13) We use: E = mc^2

and: L = hc/E for a massless particle

Then: mc^2 = hc/L

and m = h/cL

L = 6000A = 600 nm = 6 x 10^-7 m

h = 6.62 * 10^-34 J*s

c = 3 x 10^8 m/s

So:

m =3.7 x 10^-33 g

14) We have the power (1300 W) related to the E-field intensity via:

P = E^2(A)/ (2 u(o) c)

where A = 1 m^2 (area) and u(o) is the magnetic permeability = 4 pi x 10^-7 H/m

Then:

E = [P*(2 u(o) c) ]^1/2 = [1300 J/s*4 pi x 10^-7 H/m * 3 x 10^8 m/s]^1/2

= 700 V/m

15) From the Poynting vector: S = (EB)/u(o)

so: B = E/c

= (700 V/m)/ (3 x 10^8 m/s) = 2.3 x 10^-6 Wb/m^2 (weber per meter squared)

## No comments:

Post a Comment