A π0 meson is a sub-atomic particle. A stationary π0 meson, which has mass 2.4 × 10-28 kg, decays to form two γ-ray photons.

Question 9

A π₀ meson is a sub-atomic particle.

A stationary π₀ meson, which has mass 2.4 × 10^-28 kg, decays to form two γ-ray photons.

The nuclear equation for this decay is

π₀ → γ  + γ

(a) Explain why the two γ-ray photons have the same energy. [2]

(b) Determine, for each γ-ray photon,

(i) the energy, in joule, [2]

(ii) the wavelength, [2]

(iii) the momentum. [2]

Reference: Past Exam Paper – June 2010 Paper 41 Q8

Solution:

(a) As a consequence of the conservation of momentum, the momenta of the photons must be equal and opposite. Since the photons have the same momenta, they must then have the same energy.

(b)

(i)

{We need to calculate the energy of a single photon.

Mass = 1.2×10^-28 kg

Einstein’s mass-energy relation:}

(Δ)E = (Δ)mc²                                            

(Δ)E = 1.2×10^-28 × (3.0×10⁸)²

(Δ)E = 1.08×10^-11 J            

(ii)

E = hc / λ

{λ = hc / E}

λ = (6.63×10^-34 × 3.0×10⁸) / (1.08×10^-11)          

λ = 1.84×10^-14 m                                          

(iii)

{de Broglie wavelength:}

λ = h / p

p = (6.63×10^-34) / (1.84×10^-14)

p = 3.6×10^-20 N s

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.