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Date May 2018 Marks available 3 Reference code 18M.2.HL.TZ2.9
Level Higher level Paper Paper 2 Time zone 2
Command term Explain Question number 9 Adapted from N/A

Question

Rhodium-106 (10645Rh10645Rh) decays into palladium-106 (10646Pd10646Pd) 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 = nh2πh2π. 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=ke2merv=ke2mer

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=h24π2kmee2r=h24π2kmee2

[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 «mvrnh2πh2π»

[3 marks]

b.

mev2r=ke2r2mev2r=ke2r2

OR

KE = 1212PE hence 1212mev212ke2r12ke2r

«solving for v to get answer»

 

Answer given – look for correct working

[1 mark]

c.i.

combining vke2merke2mer with mevrh2πh2π using correct substitution

«eg me2ke2merr2=h24π2me2ke2merr2=h24π2»

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(6.63×1034)24π2×8.99×109×9.11×1031×(1.6×1019)2(6.63×1034)24π2×8.99×109×9.11×1031×(1.6×1019)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«hcE=6.63×1034×3×1087.68×1014 =» 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 » 12.2 – Nuclear physics
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