# 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 = p

^{2}/2m OR E_{K}= ½ mv^{2}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__x 9.0 = 72μC^{-6}
(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 = (3×10

^{8}) / 0.03 = 1×10

^{10}Hz = 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:__**(a)**A magnetic field is a region (of space) where there is a force

EITHER on / produced by magnetic
pole

OR on / produced by current carrying
conductor / moving charge

**(b)**

(i) The force on the particle is
(always) normal to the velocity / direction of travel. The speed of the particle
is constant.

(ii)

The magnetic force provides the
centripetal force.

mv

^{2}/ r = Bqv
Radius r = mv / Bq

**(c)**

(i) The direction is from ‘bottom to
top’ of the diagram

(ii) The radius is proportional to
momentum.

{r = mv / Bq. mv = p where p is the momentum. B (same field) and q
(same particle) are constant. So, the momentum p (= mv) is directly
proportional to r.}

Ratio = 5.7 / 7.4 = 0.77

__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 6.4×10

^{5}m. It completes 1 revolution in 8.6×10^{4}s. 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 / R

^{2}
(ii) F = mRω

^{2}
(iii) Reaction force = (GMm / R

^{2}) – mRω^{2}
{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 Rω

^{2}varies
OR mRω

^{2}is no longer parallel to GMm /R^{2}/ normal to surface
and becomes smaller as the object
approaches a pole / is zero at pole

{The R in the expression GMm/R

^{2}and the R in the expression mRω^{2}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 = Rω

^{2}and ω = 2πf}
Acceleration = (6.4x10

^{5}) x (2π / {8.6x10^{4}})^{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

12/O/N/10 Q.33

ReplyDelete11/M/J/11 Q.9,10

For 12/O/N/10 Q.33, go to

Deletehttp://physics-ref.blogspot.com/2014/11/9702-november-2010-paper-12-worked.html

can you please explain oct/nov 2012, variant -11, (p-1) , Q.21? :)

ReplyDeleteCheck solution 803 at

Deletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-161.html

41/M/J/11 Q.5(c)(i) Can you please show the path of the particle?

ReplyDeleteFor Q.708 Please show the path of the particle in part (c)(i)

ReplyDeleteThe path is already shown in the diagram - it's the curved line. You only need to insert the head of an arrow to indicate its direction. This is from the bottom to the top along the path.

DeleteThere could only be 2 directions - either the one stated above, or its reverse - both are along the path shown - the curved line.

Solution 705 (b)(ii)

ReplyDeleteCould you please explain the answer because I'm not sure what the question or answer is trying to specify...

By accelerating the electrons through different p.d.`s, electrons with different de Broglie wavelength are produced. Now, if one of these electrons is diffracted by a crystal, it means that the value of its de Broglie wavelength (this can be calculated as above) is similar to the separation of the atoms (this is inferred due to the diffraction taking place). Thus, diffraction allows us to predict the separation of the atoms in the crystal.

DeleteWhy when mrw^2 =0 then normal reaction =0? Isn't the normal reaction at the pole= GMm/r^2 - mRw^2? So the normal reaction will be equal to the gravitational force isn't it

ReplyDeletecould you read again. It seems that the explanation is already available above - (b)(i) in red.

Deletewhy is the acceleration 0 at the axis??

ReplyDeleteAcceleration = Rω^2 where R is the perpendicular distance from the axis. At the poles, the value of R is zero. So, the acceleration = 0

Delete