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Date May 2018 Marks available 2 Reference code 18M.2.HL.TZ2.9
Level Higher level Paper Paper 2 Time zone 2
Command term Deduce Question number 9 Adapted from N/A

Question

Rhodium-106 ( 45 106 Rh ) decays into palladium-106 ( 46 106 Pd ) by beta minus (β) decay. The diagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrow represents the β decay.

M18/4/PHYSI/HP2/ENG/TZ2/09.d

Bohr modified the Rutherford model by introducing the condition mvr = n h 2 π . Outline the reason for this modification.

[3]
b.

Show that the speed v of an electron in the hydrogen atom is related to the radius r of the orbit by the expression

v = k e 2 m e r

where k is the Coulomb constant.

[1]
c.i.

Using the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in the ground state of hydrogen is given by the following expression.

r = h 2 4 π 2 k m e e 2

[2]
c.ii.

Calculate the electron’s orbital radius in (c)(ii).

[1]
c.iii.

Explain what may be deduced about the energy of the electron in the β decay.

[3]
d.i.

Suggest why the β decay is followed by the emission of a gamma ray photon.

[1]
d.ii.

Calculate the wavelength of the gamma ray photon in (d)(ii).

[2]
d.iii.

Markscheme

the electrons accelerate and so radiate energy

they would therefore spiral into the nucleus/atoms would be unstable

electrons have discrete/only certain energy levels

the only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvrn h 2 π »

[3 marks]

b.

m e v 2 r = k e 2 r 2

OR

KE =  1 2 PE hence 1 2 mev2 1 2 k e 2 r

«solving for v to get answer»

 

Answer given – look for correct working

[1 mark]

c.i.

combining v k e 2 m e r  with mevr h 2 π  using correct substitution

«eg m e 2 k e 2 m e r r 2 = h 2 4 π 2 »

correct algebraic manipulation to gain the answer

 

Answer given – look for correct working

Do not allow a bald statement of the answer for MP2. Some further working eg cancellation of m or r must be shown

[2 marks]

c.ii.

« r ( 6.63 × 10 34 ) 2 4 π 2 × 8.99 × 10 9 × 9.11 × 10 31 × ( 1.6 × 10 19 ) 2 »

r = 5.3 × 10–11 «m»

[1 mark]

c.iii.

the energy released is 3.54 – 0.48 = 3.06 «MeV»

this is shared by the electron and the antineutrino

so the electron’s energy varies from 0 to 3.06 «MeV»

[3 marks]

d.i.

the palladium nucleus emits the photon when it decays into the ground state «from the excited state»

[1 mark]

d.ii.

Photon energy

E = 0.48 × 106 × 1.6 × 10–19«7.68 × 10–14 J»

λ« h c E = 6.63 × 10 34 × 3 × 10 8 7.68 × 10 14 =» 2.6 × 10–12 «m»

 

Award [2] for a bald correct answer

Allow ECF from incorrect energy

[2 marks]

d.iii.

Examiners report

[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
c.iii.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
d.iii.

Syllabus sections

Additional higher level (AHL) » Topic 12: Quantum and nuclear physics » 12.1 – The interaction of matter with radiation
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Additional higher level (AHL) » Topic 12: Quantum and nuclear physics
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