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: a = a_x(i^) + a_y(j^) + a_z(k^)
and - as before: r^ = x(i^) + y(j^) = z(k^)
Then: a*r = a_x(x) + a_y(y) + a_z(z)
Then the gradient, grad (a*r) =
i^ @(a_x(x))/@x + j^2 @(a_y(y))/@y + k^ @(a_z(z))@z
Then, grad (a*r) = 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)
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 = a_x(i^) + a_y(j^) + a_z(k^)
and - as before: r^ = x(i^) + y(j^) = z(k^)
Then: a*r = a_x(x) + a_y(y) + a_z(z)
Then the gradient, grad (a*r) =
i^ @(a_x(x))/@x + j^2 @(a_y(y))/@y + k^ @(a_z(z))@z
Then, grad (a*r) = 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)
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