Solutions To GRE Physics Problems (1)

 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, and f is the frequency of the light.

l1 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/l1]

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 10³  eV* Å

then,

E = 12.4 x 10³ eV*Å [ 1/1800Å - 1/ 2300Å] = 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 π² me⁴/h²)(Z² /n²)

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

E(n) = - 121(2 π² me⁴ /h²)

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)


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^  ¶(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 will 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.

l²(1 - l) - (1 - l) = 0, or

(1 - l) (l² - 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:

N² (U1 + 2i U2)(U1* - 2i (U2*) = 1


Therefore:

5N² = 1

So: N = (5)^-1/2


7) The sketch for the problem is shown here:

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) d² (i^) /dt² = 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 below, with the Lagrangian (L) boxed:

12) We use the relativistic form:

E= mc²[1 - (v/c)²]^-1/2 - 1]

Noting the quickie conversion factor: mc² = 0.511 MeV

Now, if E = 0.25 MeV, then:

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

and:

3/2 = [1 - (v/c)²]^-1/2

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)^1/2 = 0.75

so, v = 0.75c


13) We use: E = mc²

and: l = hc/E for a massless particle

Then: mc² = 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 10⁸ m/s

So:

m =3.7 x 10^-33 g