Question 18M.2.HL.TZ2.9

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 «mvr = n$\frac{h}{2\pi }$»

[3 marks]

$\frac{m_{\text{e}}{v}^2}{r}=\frac{k{e}^2}{r^2}$

OR

KE = $\frac{1}{2}$PE hence $\frac{1}{2}$m_ev² = $\frac{1}{2}\frac{k{e}^2}{r}$

«solving for v to get answer»

Answer given – look for correct working

[1 mark]

combining v = $\sqrt{\frac{k{e}^2}{m_{\text{e}}r}}$ with m_evr = $\frac{h}{2\pi }$ using correct substitution

«eg ${m_e}^2\frac{k{e}^2}{m_{\text{e}}r}{r}^2=\frac{h^2}{4{\pi }^2}$»

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]

« r = $\frac{{(6.63\times {10}^{-34})}^2}{4{\pi }^2\times 8.99\times {10}^9\times 9.11\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$»

r = 5.3 × 10^–11 «m»

[1 mark]

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]

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

[1 mark]

Photon energy

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

λ = «$\frac{hc}{E}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{7.68\times {10}^{-14}}$ =» 2.6 × 10^–12 «m»

Award [2] for a bald correct answer

Allow ECF from incorrect energy

[2 marks]