IB Physics HL D 2 12 Hl Two Charge Potential Energy QuestionsDerive two-charge electric potential energy and total orbital energy, then interpret how energy loss changes an electron’s orbit and wavelength.SyllabusFirst assessment 2025CoursePhysics HLLevelHL
Exam pointsDerive and use the two-charge electric potential energy relation Ep=kq1q2/r, including sign.Relate changes in two-charge potential energy to total energy and orbital radius.
IB Physics HL D 2 12 Hl Two Charge Potential Energy Questions question 1[Maximum number: 4]Question (a)(a)In a classical model of the singly-ionized helium atom, a single electron orbits the nucleus in a circular orbit of radius r.[ 4 ]Question (i)(i)Hence, deduce that the total energy of the electron is given by ETOT=−ke2rE_{\mathrm{TOT}}=-\frac{k e^{2}}{r}ETOT=−rke2.[ 2 ]Show Answeruses (a)(i) to state Ek=ke2rORE_{\mathrm{k}}=\frac{k e^{2}}{r} \mathcal{OR}Ek=rke2OR states Ep=−2ke2rE_{\mathrm{p}}=-\frac{2 k e^{2}}{r}Ep=−r2ke2 adds «ETOT=Ek+Ep=ke2r−2ke2r« E_{\mathrm{TOT}}=E_{\mathrm{k}}+E_{\mathrm{p}}=\frac{k e^{2}}{r}-\frac{2 k e^{2}}{r}«ETOT=Ek+Ep=rke2−r2ke2 » to get the result Question (ii)(ii)In this model the electron loses energy by emitting electromagnetic waves.Describe the predicted effect of this emission on the orbital radius of the electron.[ 2 ]Show Answerthe total energy decreases OR by reference to ETOT =−ke2rE_{\text {TOT }}=-\frac{k e^{2}}{r}ETOT =−rke2 the radius must also decreaseMarking guidance:Award [0] for an answer concluding that radius increases8.biwith n=3, v= « 2×8.99×109×(1.6×10−19)29.11×10−31×9×2.7×10−11=\sqrt{\frac{2 \times 8.99 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{9.11 \times 10^{-31} \times 9 \times 2.7 \times 10^{-11}}}=9.11×10−31×9×2.7×10−112×8.99×109×(1.6×10−19)2= » 1.44×1061.44 \times 10^{6}1.44×106 « ms−1\mathrm{ms}^{-1}ms−1 »λ=6.63×10−349.11×10−31×1.44×106 OR λ=5.05×10−10<m≫\lambda=\frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.44 \times 10^{6}} \text { OR } \lambda=5.05 \times 10^{-10}<\mathrm{m} \ggλ=9.11×10−31×1.44×1066.63×10−34 OR λ=5.05×10−10<m≫Add to Test
Question (a)(a)In a classical model of the singly-ionized helium atom, a single electron orbits the nucleus in a circular orbit of radius r.[ 4 ]Question (i)(i)Hence, deduce that the total energy of the electron is given by ETOT=−ke2rE_{\mathrm{TOT}}=-\frac{k e^{2}}{r}ETOT=−rke2.[ 2 ]Show Answeruses (a)(i) to state Ek=ke2rORE_{\mathrm{k}}=\frac{k e^{2}}{r} \mathcal{OR}Ek=rke2OR states Ep=−2ke2rE_{\mathrm{p}}=-\frac{2 k e^{2}}{r}Ep=−r2ke2 adds «ETOT=Ek+Ep=ke2r−2ke2r« E_{\mathrm{TOT}}=E_{\mathrm{k}}+E_{\mathrm{p}}=\frac{k e^{2}}{r}-\frac{2 k e^{2}}{r}«ETOT=Ek+Ep=rke2−r2ke2 » to get the result Question (ii)(ii)In this model the electron loses energy by emitting electromagnetic waves.Describe the predicted effect of this emission on the orbital radius of the electron.[ 2 ]Show Answerthe total energy decreases OR by reference to ETOT =−ke2rE_{\text {TOT }}=-\frac{k e^{2}}{r}ETOT =−rke2 the radius must also decreaseMarking guidance:Award [0] for an answer concluding that radius increases8.biwith n=3, v= « 2×8.99×109×(1.6×10−19)29.11×10−31×9×2.7×10−11=\sqrt{\frac{2 \times 8.99 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{9.11 \times 10^{-31} \times 9 \times 2.7 \times 10^{-11}}}=9.11×10−31×9×2.7×10−112×8.99×109×(1.6×10−19)2= » 1.44×1061.44 \times 10^{6}1.44×106 « ms−1\mathrm{ms}^{-1}ms−1 »λ=6.63×10−349.11×10−31×1.44×106 OR λ=5.05×10−10<m≫\lambda=\frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.44 \times 10^{6}} \text { OR } \lambda=5.05 \times 10^{-10}<\mathrm{m} \ggλ=9.11×10−31×1.44×1066.63×10−34 OR λ=5.05×10−10<m≫
Question (i)(i)Hence, deduce that the total energy of the electron is given by ETOT=−ke2rE_{\mathrm{TOT}}=-\frac{k e^{2}}{r}ETOT=−rke2.[ 2 ]Show Answeruses (a)(i) to state Ek=ke2rORE_{\mathrm{k}}=\frac{k e^{2}}{r} \mathcal{OR}Ek=rke2OR states Ep=−2ke2rE_{\mathrm{p}}=-\frac{2 k e^{2}}{r}Ep=−r2ke2 adds «ETOT=Ek+Ep=ke2r−2ke2r« E_{\mathrm{TOT}}=E_{\mathrm{k}}+E_{\mathrm{p}}=\frac{k e^{2}}{r}-\frac{2 k e^{2}}{r}«ETOT=Ek+Ep=rke2−r2ke2 » to get the result
Question (ii)(ii)In this model the electron loses energy by emitting electromagnetic waves.Describe the predicted effect of this emission on the orbital radius of the electron.[ 2 ]Show Answerthe total energy decreases OR by reference to ETOT =−ke2rE_{\text {TOT }}=-\frac{k e^{2}}{r}ETOT =−rke2 the radius must also decreaseMarking guidance:Award [0] for an answer concluding that radius increases8.biwith n=3, v= « 2×8.99×109×(1.6×10−19)29.11×10−31×9×2.7×10−11=\sqrt{\frac{2 \times 8.99 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{9.11 \times 10^{-31} \times 9 \times 2.7 \times 10^{-11}}}=9.11×10−31×9×2.7×10−112×8.99×109×(1.6×10−19)2= » 1.44×1061.44 \times 10^{6}1.44×106 « ms−1\mathrm{ms}^{-1}ms−1 »λ=6.63×10−349.11×10−31×1.44×106 OR λ=5.05×10−10<m≫\lambda=\frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.44 \times 10^{6}} \text { OR } \lambda=5.05 \times 10^{-10}<\mathrm{m} \ggλ=9.11×10−31×1.44×1066.63×10−34 OR λ=5.05×10−10<m≫