** 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 / 4****p ExB **

S = direction of the electromagnetic wave

E = Electric field

B = Magnetic field

c) **X-rays has the smallest wavelength (10 ^{-11} – 10^{-8}m) **

** Radio waves has the largest wavelength (>10 ^{-1}m) **

** **

**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 10 ^{8} / 1.225 = 2.4489 x 10^{8} 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 10^{15} /4.13566 = **2.926 x 10 ^{15} Hz**

** **

b) The equation is En = E /n^{2 }

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 10^{8} 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 10^{8} / 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 cm ^{2} 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 cm^{2} = 400 x 10^{-4} m^{2}

Dt = 0.01s

q= 0

** EMF = 100 x 0.1 x 4 x10 ^{-2} x1 / 0.01 = 0.4 volts**