An X-ray photon of energy 3.06 × 10-14 J is incident on an isolated stationary electron, as illustrated in Fig. 6.1.

Question 8

(a) Explain what is meant by a photon. [2]

(b) An X-ray photon of energy 3.06 × 10^-14 J is incident on an isolated stationary electron, as illustrated in Fig. 6.1.

Fig. 6.1

The photon is deflected elastically by the electron through angle θ. The deflected photon has a wavelength of 6.80 × 10^-12 m.

(i) On Fig. 6.1, draw an arrow to indicate a possible initial direction of motion of the electron after the photon has been deflected. [1]

(ii) Calculate

1. the energy of the deflected photon, [2]

2. the speed of the electron after the photon has been deflected. [3]

(c) Explain why the magnitude of the final momentum of the electron is not equal to the change in magnitude of the momentum of the photon. [2]

Reference: Past Exam Paper – June 2015 Paper 42 Q6

Solution:

(a) A photon is a discrete amount (packet/quantum) of electromagnetic energy.

(b)

(i) arrow below axis and pointing to right

{Momentum before collision = Momentum after collision

Initially, there is no vertical component of momentum.

So, the resultant vertical component of momentum after collision should also be zero.

The deflected photon has an upward vertical component.

So, the electron should have a downward vertical component so that the 2 components cancel each other.}

(ii)

1.

{Energy of photon: E = hf = hc / λ }

E = hc / λ

E = (6.63×10^-34 × 3.0×10⁸) / (6.80×10^-12)       

E = 2.93×10^-14 J

2.

{Energy lost by photon = energy gained by electron

Energy lost be photon = (3.06 – 2.93) × 10^-14 J}

energy of electron = (3.06 – 2.93) × 10^-14

energy of electron = 1.3 × 10^-15 J

{Energy E = ½ mv²

Speed v = √(2E/m)}

speed = √(2E/m)        

speed = 5.4 × 10⁷ m s^-1             

(c) Momentum is a vector quantity. So we need to consider the changes in direction, not just the magnitude. We must consider the momentum in two directions.

The diagrams show a metal cube suspended from a spring balance before and during immersion in water.

Question 35

The diagrams show a metal cube suspended from a spring balance before and during immersion in water.

A reduction in the balance reading occurs as a consequence of the immersion.

Which statement is correct?

A The balance reading will be further reduced if the cube is lowered further into the water.

B The balance reading during immersion corresponds to the upthrust of the water on the cube.

C The forces acting on the vertical sides of the cube contribute to the change in the balance reading.

D The gravitational pull on the cube is unchanged by the immersion.

Reference: Past Exam Paper – November 2010 Paper 12 Q13

Solution:

Answer: D.

Before the immersion, the balance reads the weight of the object, which is downwards. The weight of the cube is the gravitational pull on the cube and this is unchanged in a specific location since weight W = mg. [D is correct]

As the cube is immersed, it experiences an upthrust force upwards. This force is due to the difference in pressures acting on the top and bottom of the cube.

Pressure = hρg

The pressure on the bottom face of the cube is greater than the pressure at the top face. This is because the lower face is at a greater depth in the water.

Pressure = Δhρg

where Δh is the difference in depth of the top and bottom faces.

It is be noted that Δh is also the length of the cube.

For whatever depth of immersion of the cube, the pressure is given by P = Δhρg. Since the length of the cube does not change (Δh is constant), the upthrust force on the cube will always have the same value at any depth when fully immersed. Thus, the upthrust is unchanged no matter how further the cube is immersed. [A is incorrect]

The forces acting on the vertical sides of the cube are equal in magnitude since they act at the same depth in the water. So, they cancel each other out and are not the cause in the change in the balance reading. [C is incorrect] The balance reading is affected by the vertical forces acting on the cube (weight and upthrust).

Thus, the balance reading is not the upthrust but the weight (which acts downwards) minus the upthrust (which acts upwards). The balance reads the resultant of these two forces. [B is incorrect]

A loudspeaker produces a sound wave of constant frequency. Outline how a cathode-ray oscilloscope (c.r.o.) may be used to determine this frequency.

Question 20

A loudspeaker produces a sound wave of constant frequency.

Outline how a cathode-ray oscilloscope (c.r.o.) may be used to determine this frequency. [4]

Reference: Past Exam Paper – November 2010 Paper 23 Q3

Solution:

A microphone (or terminals of a loudspeaker) is connected to the Y-plates of the c.r.o.     

The c.r.o. is adjusted to produce a steady wave of 1 (or 2) cycles (or wavelengths) on screen {using the time-base setting}.

The length of a cycle (wavelength λ) is measured and the time-base b is noted.

{Period T = length of cycle × time-base = λb

Frequency = 1 / T = 1 / λb }

frequency = 1 / λb