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 10

^{3}eV* Å

then,

E = 12.4 x 10

^{3}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 π

^{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 π

^{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)

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:

***r = a**

*a*_{x}(x) + a

_{y}(y) + a

_{z}(z)

Then the gradient, grad (

***r) =**

*a***i^**¶ (a

_{x}(x))/ ¶x +

**j**^ ¶(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 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.

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)

^{-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: d

**i**^/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 below, with the Lagrangian (L) boxed:

12) We use the relativistic form:

E= mc

^{2}[1 - (v/c)

^{2}]

^{-1/2}- 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}]

^{-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 = 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 m

_{o}c)

where A = 1 m

^{2}(area) and m

_{o }is the magnetic permeability = 4 π x 10

^{-7}H/m

Then:

E = [P*(2 m

_{o}c) ]

^{1/2}= [1300 J/s*4 π x 10

^{-7}H/m * 3 x 10

^{8}m/s]

^{1/2}

= 700 V/m

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

_{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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