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A-Level CAIE Physics 22 1 Energy And Momentum Of A Photon Question Bank

Practice A-Level CAIE Physics 22 1 Energy And Momentum Of A Photon questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.

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Question 6

6

7 marks

Question 6(a)

6(a)

Explain what is meant by a photon.

Easystructured2 marks

Answer

packet/quantum/discrete amount of energy of electromagnetic energy/radiation/waves A1

Question 6(b)

6(b)

An X-ray photon of energy \(3.06 \times 10^{-14} \mathrm{~J}\) is incident on an isolated stationary electron, as illustrated in Fig. 6.1. The photon is deflected elastically by the electron through angle \(\theta\). The deflected photon has a wavelength of \(6.80 \times 10^{-12} \mathrm{~m}\).

structured3 marks

Question 6(b)(ii)

6(b)(ii)

Calculate 1. the energy of the deflected photon, 2. the speed of the electron after the photon has been deflected. \(\mathrm{m} \mathrm{s}^{-1}[3]\)

Hardstructured3 marks

Answer

1. \(E=h c / \lambda\) 2. energy of electron \(=(3.06-2.93) \times 10^{-14}\) speed \(=\sqrt{(2 E / m)}\)

Question 6(c)

6(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.

Hardstructured2 marks

Answer

momentum is a vector quantity either must consider momentum in two directions or direction changes so cannot just consider magnitude

Question 8

8

White light is incident on a cloud of cool hydrogen gas, as illustrated in Fig. 8.1. The spectrum of the light emerging from the gas cloud is found to contain a number of dark lines.

structured4 marks

Question 8(b)

8(b)

Some electron energy levels in a hydrogen atom are illustrated in Fig. 8.2. One dark line is observed at a wavelength of 435 nm .

structured4 marks

Question 8(b)(i)

8(b)(i)

Calculate the energy, in eV , of a photon of light of wavelength 435 nm .

Mediumstructured4 marks

Answer

\(E=h c / \lambda\) C1 \(\begin{array}{ll}=\left(6.63 \times 10^{-34} \times 3 \times 10^{8}\right) /\left(435 \times 10^{-9}\right) & \mathrm{C} 1 \\ =4.57 \times 10^{-19} \mathrm{~J} \text { (allow } 2 \text { s.f.) } & \mathrm{C} 1\end{array}\) \(=\left(4.57 \times 10^{-19}\right) /\left(1.6 \times 10^{-19}\right)(\mathrm{eV})\) =2.86 eV (allow 2 s.f.) A1 [4]

Question 8

8

Light of wavelength 590 nm is incident normally on a surface, as illustrated in Fig. 8.1. The power of the light is 3.2 mW . The light is completely absorbed by the surface.

structured4 marks

Question 8(a)

8(a)

Calculate the number of photons incident on the surface in 1.0 s .

Mediumstructured0 marks

Answer

photon energy \(=h c / \lambda\)

Question 8(b)

8(b)

Use your answer in (a) to determine

structured4 marks

Question 8(b)(i)

8(b)(i)

the total momentum of the photons arriving at the surface in 1.0 s , \(\mathrm{kgms}^{-1}\)

Mediumstructured3 marks

Answer

\(p=h / \lambda \quad \mathrm{C} 1\)

Question 8(b)(ii)

8(b)(ii)

the force exerted on the surface by the light.

Mediumstructured1 marks

Answer

force \(=1.06 \times 10^{-11} \mathrm{~N}\)

Question 6

6

3 marks

Question 6(c)

6(c)

The \(\beta\)-particle is emitted with an energy of \(5.7 \times 10^{3} \mathrm{eV}\). Calculate the speed of the \(\beta\)-particle. speed = \(\mathrm{ms}^{-1}\)

Hardstructured3 marks

Answer

energy \(=5.7 \times 10^{3} \times 1.6 \times 10^{-19}\left(=9.12 \times 10^{-16} \mathrm{~J}\right)\) ..... C1 \(v^{2}=\frac{2 \times 9.12 \times 10^{-16}}{9.11 \times 10^{-31}}\) ..... C1 \(v=4.5 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}\) ..... A1 [3]

Question 8

8

3 marks

Question 8(a)

8(a)

State what is meant by a photon.

Easystructured2 marks

Answer

discrete amount/packet/quantum of energy M1 of electromagnetic radiation/EM radiation A1 [2]

Question 8(b)

8(b)

A beam of light is incident normally on a metal surface, as illustrated in Fig. 8.1. The beam of light has cross-sectional area \(1.3 \times 10^{-5} \mathrm{~m}^{2}\) and power \(2.7 \times 10^{-3} \mathrm{~W}\). The light has wavelength 570 nm . The light energy is absorbed by the metal and no light is reflected.

structured1 marks

Question 8(b)(i)

8(b)(i)

Show that a photon of this light has an energy of \(3.5 \times 10^{-19} \mathrm{~J}\).

Mediumstructured1 marks

Answer

\(E=h c / \lambda\)

Question 8(b)(ii)

8(b)(ii)

Calculate, for a time of 1.0 s , 1. the number of photons incident on the surface, number = 2. the change in momentum of the photons. change in momentum = \(\mathrm{kg} \mathrm{ms}^{-1}\)

Hardstructured0 marks

Answer

1. number \(=\left(2.7 \times 10^{-3}\right) /\left(3.5 \times 10^{-19}\right)\) 2. momentum of photon \(=h / \lambda\) (allow E=p c route to \(9 \times 10^{-12}\) )

Question 8

8

3 marks

Question 8(a)

8(a)

Explain what is meant by a photon.

Easystructured3 marks

Answer

packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 (allow 1 mark for 'packet of electromagnetic radiation') energy = Planck constant × frequency (seen here or in b )

Question 8

8

A photon of wavelength \(6.50 \times 10^{-12} \mathrm{~m}\) is incident on an isolated stationary electron, as illustrated in Fig. 8.1. The photon is deflected elastically by the electron of mass \(m_{\mathrm{e}}\). The wavelength of the deflected photon is \(6.84 \times 10^{-12} \mathrm{~m}\).

structured7 marks

Question 8(a)

8(a)

Calculate, for the incident photon,

structured2 marks

Question 8(a)(i)

8(a)(i)

its momentum, momentum = Ns

Easystructured2 marks

Answer

\(p=h / \lambda\)

Question 8(a)(ii)

8(a)(ii)

its energy. energy =

Easystructured0 marks

Answer

\(E=h c / \lambda\) or E=p c

Question 8(b)

8(b)

The angle \(\theta\) through which the photon is deflected is given by the expression where \(\Delta \lambda\) is the change in wavelength of the photon, h is the Planck constant and c is the speed of light in free space.

structured5 marks

Question 8(b)(i)

8(b)(i)

Calculate the angle \(\theta\).

Hardstructured2 marks

Answer

\(0.34 \times 10^{-12}=\left(6.63 \times 10^{-34}\right) /\left(9.11 \times 10^{-31} \times 3.0 \times 10^{8}\right) \times(1-\cos \theta) \quad\) C1

Question 8(b)(ii)

8(b)(ii)

Use energy considerations to suggest why \(\Delta \lambda\) must always be positive.

Hardstructured3 marks

Answer

deflected electron has energy ..... M1 this energy is derived from the incident photon ..... A1 deflected photon has less energy, longer wavelength (so \(\Delta \lambda\) always positive) ..... B1 [3]

Question 8

8

The power for a space probe is to be supplied by the energy released when plutonium- 236 decays by the emission of \(\alpha\)-particles. The \(\alpha\)-particles, each of energy 5.75 MeV , are captured and their energy is converted into electrical energy with an efficiency of 24 %.

structured1 marks

Question 8(a)

8(a)

Calculate

structured1 marks

Question 8(a)(i)

8(a)(i)

the energy, in joules, equal to 5.75 MeV , energy = J

Easystructured1 marks

Answer

energy \(=5.75 \times 1.6 \times 10^{-13}\)

Question 7

7

6 marks

Question 7(a)

7(a)

A photon has an energy of \(3.11 \times 10^{-19} \mathrm{~J}\). Calculate the momentum of the photon.

Easystructured2 marks

Answer

p=E / c C1 \[ =\left(3.11 \times 10^{-19}\right) /\left(3.00 \times 10^{8}\right) \] \(=1.04 \times 10^{-27} \mathrm{~N} \mathrm{~s}\) A1

Question 7(b)

7(b)

A laser beam has a power of 350 mW . The light from the laser has a wavelength of 640 nm .

structured4 marks

Question 7(b)(i)

7(b)(i)

Determine the number of photons emitted by the laser in a time of 1.0 s .

Mediumstructured2 marks

Answer

E=h f and \(c=f \lambda\) so energy of one photon \(=h c / \lambda\) \[ 350 \times 10^{-3}=N \times\left(6.63 \times 10^{-34} \times 3.00 \times 10^{8}\right) /\left(640 \times 10^{-9}\right) \] C1 \(N=1.1 \times 10^{18}\) A1

Question 7(b)(ii)

7(b)(ii)

The laser beam is incident normally on a surface that absorbs all of the photons. Show that the force F exerted on the surface by the laser beam is given by where P is the power of the laser beam and c is the speed of light.

Mediumstructured2 marks

Answer

F= (change in) momentum / time M1 Clear use of p=E / c and t=E / P to complete the algebra and arrive at the final equation: \[ \text { e.g. } F=[E / c] /[E / P]=P / c \] A1

Question 8

8

\(8 \mathrm{~A} \pi^{0}\) meson is a sub-atomic particle. A stationary \(\pi^{0}\) meson, which has mass \(2.4 \times 10^{-28} \mathrm{~kg}\), decays to form two \(\gamma\)-ray photons. The nuclear equation for this decay is

structured2 marks

Question 8(b)

8(b)

Determine, for each \(\gamma\)-ray photon,

structured2 marks

Question 8(b)(i)

8(b)(i)

the energy, in joule, energy = J

Mediumstructured2 marks

Answer

\(\begin{array}{rlr}(\Delta) E & =(\Delta) m c^{2} & \text { C1 } \\ & =1.2 \times 10^{-28} \times\left(3.0 \times 10^{8}\right)^{2} & \end{array}\)

Question 8(b)(iii)

8(b)(iii)

the momentum. Answer all the questions in the spaces provided.

Mediumstructured0 marks

Answer

\(\lambda=h / p\) \(p=\left(6.63 \times 10^{-34}\right) /\left(1.84 \times 10^{-14}\right) \quad \mathrm{C} 1\) C1 C1 A1 8 M1 A1 [2] 1 A1 Section B