Physics worked examples on force,wavelength,energy

1.  Can a charged particle move through a magnetic field without experiencing any force? How?

b.  By measuring the electric and magnetic field at a point in space where there is an electromagnetic wave, can one determine the direction of the wave? How?

c.  Consider light, xrays, radio waves. Which of these has the smallest wavelength? And the largest?

Solution:

A)  A charge can move in a uniform magnetic without experiencing any force.

F = q (VxB) = qVB sinq

When the angle is either 0 or 180 between the V and B, the force acting on the charge particle is Zero.

F = qVB sin0 = 0, F = qVBsin180 = 0

b) Yes, the direction of the Electromagnetic wave can be determined by calculating the electric and magnetic field at a point in space by using pointing vector.

S = c / 4p ExB

S = direction of the electromagnetic wave

E = Electric field

B = Magnetic field

c) X-rays has the smallest wavelength (10-11 – 10-8m)

Radio waves has the largest wavelength (>10-1m)

2.  A ray of light strikes a block of some transparent material at an angle of incidence of 60degrees.

a.  If the angle  of refraction is 45degrees, determine the index of refraction of the material, assuming the ray travels in a vacuum before striking the block.

b. Find the speed of light in the block.

Solution:

A) The refractive index  formula

Sini/sinr = V1/v2 = n2/n1

I= angle of incidence = 60

R= angle of refraction = 45

N2= refractive index of second material

N1 = refractive index of the incidence material = 1

N= sin60/sin45 = 0.866/0.707 = 1.2249 = 1.225

b) Speed of light of the block (V2)

V1/V2 = n= 1.225

V2 = V1/1.225 = 3 x 108 / 1.225 = 2.4489 x 108 m/s

3.  A hydrogen atom emits a photon having an energy of 12.1 eV.

b.  What’s the frequency of the photon?

c.  Using the fact that the ionization energy of the H-atom in its ground state is 13.6 eV, determine the principal quantum number “n” of the initial state, given that for the final state, “n” = 1.

Solution :

E =  h f

E = energy of the photon

F = frequency

E = 12.1eV

H= 4.13566 x 10-15 eV.s

12.1 =4.13566 x10-15 f

F = 12.1 x 1015 /4.13566 = 2.926 x 1015 Hz

b) The equation is En = E /n2

The principal quantum number range from n= 1, 2, 3………

En = -13.6/1 = -13.6

So the principal quantum numbers are n= 1, 2, 3……….

4.  Electrons with a maximum energy of 2.8 eV are emitted from a surface radiated with light of wavelength 200 nm

a.  Find energy in eV of the incident light

b.  Find the smallest energy which the light could have and still free an electron from the surface of the metal.

Solution:

E = hc/l

C = 3 x 108 m/s

H= 4.13566 x 10-15 eV.s

l = 200 x 10-9 m

E = 4.13566 x 10-15 x 3 x 108 / 200 x 10-9 = 0.06203 x 100= 6.203 eV

b) The maximum energy of the electron = 2.8ev

Maximum kinetic energy = Energy of the incident photon – minimum energy required to remove a delocalized electron its surface (W)

2.8 = 6.203 – W

W = 6.203 -2.8 = 3.403 eV (minimum energy required)

5.  A coil of 100 turns, with a cross sectional area of 400 cm2 is placed with its plane perpendicular to a uniform magnetic field of .1 T and its rotated through a quarter turn (90 degrees) in .01 so that its plane becomes parallel to the field.

What is the average induced emf?

Solution:

Induced EMF = Nd(BA cosq)dt

N = number of turns = 100

B = 0.1 T

A = 400 cm2 = 400 x 10-4 m2

Dt = 0.01s

q= 0

EMF = 100 x 0.1 x 4 x10-2 x1 / 0.01 = 0.4 volts