• Physics 9702 Doubts | Help Page 143

Question 705: [Quantum Physics]

(a) State what is meant by de Broglie wavelength.

(b) An electron is accelerated from rest in a vacuum through potential difference of 4.7 kV.

(i) Calculate de Broglie wavelength of the accelerated electron.

(ii) By reference to answer in (i), suggest why such electrons may assist with an understanding of crystal structure.

Reference: Past Exam Paper – November 2012 Paper 43 Q7

Solution 705:

(a) The de Broglie wavelength is the wavelength associated with a particle that is moving

(b)

(i)

Kinetic energy {= VQ} = 1.6 × 10^–19 × 4700 = 7.52 × 10^–16 J

EITHER energy = p²/2m        OR E_K = ½ mv² and p = mv

Momentum p = √(7.52×10^–16 × 2 × 9.1×10^–31) = 3.7 × 10^–23 N s

Wavelength λ = h / p = (6.63 × 10^–34) / (3.7 × 10^–23) = 1.8 × 10^–11 m

                          

(ii) The wavelength is about the separation of atoms, so they can be used in (electron) diffraction.

 {The separation of atoms can be identified by diffraction of the electrons.}

Question 706: [Current of Electricity > Capacitance]

(a)

(i) Define capacitance.

(ii) A capacitor is made of two metal plates, insulated from one another, as shown in Fig.1.

Explain why capacitor is said to store energy but not charge.

(b) Three uncharged capacitors X, Y and Z, each of capacitance 12 μF, are connected as shown in Fig.2.

Potential difference of 9.0 V is applied between points A and B.

(i) Calculate combined capacitance of the capacitors X, Y and Z.

(ii) Explain why, when potential difference of 9.0 V is applied, the charge on one plate of capacitor X is 72 μC.

(iii) Determine

1. potential difference across capacitor X

2. charge on one plate of capacitor Y

Reference: Past Exam Paper – November 2012 Paper 41 & 42 Q5

Solution 706:

(a)

(i) Capacitance is the ratio of charge and potential (difference) / voltage.

(ii)  On each of the two plates, the capacitor has equal magnitudes of (+)ve and (–)ve charge. The total charge on the capacitor is therefore zero, and so, the capacitor does not store charge. Work needs to be done to separate the (+)ve and (-)ve charges on the plates. So, the capacitor stores energy to achieve this.

(b)

(i)

Capacitance of Y and Z together is 24μF.

1/C = (1/24) + (1/12)

C = 8.0μF 

(ii) (Some discussion as to why all charge of one sign on one plate of X.)

As the p.d. is applied, the positive charge would flow from positive terminal to each one plate of the capacitor, which would then become positively charged.

Q = CV = 8.0x10^-6 x 9.0 = 72μC

(iii) Determine

1. V = Q / C = (8.0x10^-6) / (12x10^-6) = 6.0V

2.

Either Q = CV = 12x10^-6 x 3.0 or charge is shared between Y and Z (9V – 6V)

Charge = 36μC.

Question 707: [Waves]

Microwave transmitter emits waves towards a metal plate. Waves strike the plate and are reflected back along their original path.

A microwave detector is moved along line PT.

Points P, Q, R, S and T are the positions where minima of intensity are observed. These points are found to be 15 mm apart.

What is the frequency of the microwaves?

A 5.0 GHz                  B 6.7 GHz                   C 10 GHz                    D 20 GHz

Reference: Past Exam Paper – November 2010 Paper 12 Q26

Solution 707:

Answer: C.
A stationary wave is formed. The distance between 2 consecutive minima gives half the wavelength.
Wavelength λ = 2(15) = 30mm = 0.03m

Speed v = f λ
Frequency, f = v / λ = (3×10⁸) / 0.03 = 1×10¹⁰Hz = 10GHz

Question 708: [Electromagnetism]

(a) State what is meant by a magnetic field.

(b) Charged particle of mass m and charge +q is travelling with velocity v in a vacuum. It enters a region of uniform magnetic field of flux density B, as shown in Fig.1.

Magnetic field is normal to the direction of motion of the particle. Path of the particle in the field is the arc of a circle of radius r.

(i) Explain why path of the particle in the field is the arc of a circle.

(ii) Show that radius r is given by the expression

r = mv / Bq

(c) A thin metal foil is placed in magnetic field in (b).

A second charged particle enters region of the magnetic field. It loses kinetic energy as it passes through the foil. The particle follows the path shown in Fig.2.

(i) On Fig.2, mark with an arrow the direction of travel of the particle.

(ii) Path of the particle has different radii on each side of the foil.

The radii are 7.4 cm and 5.7 cm.

Determine the ratio

final momentum of particle / initial momentum of particle

for the particle as it passes through the foil.

Reference: Past Exam Paper – June 2011 Paper 41 Q5

Solution 708:

Go to
A charged particle of mass m and charge +q is travelling with velocity v in a vacuum.  It enters a region of uniform magnetic field of flux density B, as shown in Fig. 5.1.

Question 709: [Gravitation]

Spherical planet has mass M and radius R.

The planet may be assumed to be isolated in space and to have its mass concentrated at centre.

Planet spins on its axis with angular speed ω, as illustrated in Fig.1.

Small object of mass m rests on equator of the planet. Surface of planet exerts a normal reaction force on the mass.

(a) State formulae, in terms of M, m, R and ω for

(i) gravitational force between planet and object

(ii) centripetal force required for circular motion of the small mass

(iii) normal reaction exerted by planet on the mass

(b)

(i) Explain why normal reaction on the mass will have different values at equator and at the poles

(ii) Radius of planet is 6.4×10⁵m. It completes 1 revolution in 8.6×10⁴s. Calculate magnitude of centripetal acceleration at

1. equator

2. one of the poles

(c) Suggest 2 factors that could, in case of a real planet, cause variations in the acceleration of free fall at its surface

Reference: Past Exam Paper – November 2008 Paper 4 Q1

Solution 709:

(a)

(i) F = GMm / R²

(ii) F = mRω²

(iii) Reaction force = (GMm / R²) – mRω²

{The force of gravity is attractive, so it attracts the mass towards the planet. The normal reaction opposes the gravitational force and thus acts outwards, away from the planet. Now, the resultant of these 2 forces provides the centripetal force.

Gravitational force – Normal reaction = Centripetal force

Normal reaction = Gravitational force – Centripetal force}

(b)

(i)

EITHER The value of R in the expression Rω² varies          

OR mRω² is no longer parallel to GMm /R² / normal to surface

and becomes smaller as the object approaches a pole / is zero at pole

{The R in the expression GMm/R² and the R in the expression mRω² are different. In the first one, R is the radius of the planet (considered to be spherical). In the second equation, R is the perpendicular distance of the mass from the axis of spin of the planet. This axis passes through the centre of the sphere. At the equator, R is the same in both cases. However, at the poles, the ‘perpendicular distance of the mass from the axis of spin of the planet’ is actually zero. At the top of the sphere (pole), the distance of the surface from the axis is zero.}

(ii)

1.

{Acceleration = Rω² and ω = 2πf}

Acceleration = (6.4x10⁵) x (2π / {8.6x10⁴})² = 0.034ms^-2

2. Acceleration = 0

(c) Choose any 2:

The ‘radius’ of the planet varies

The density of the planet is not constant

The spinning of the planet

Nearby planets / stars