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 pointscalculate electric potential energy and total orbital energy for an electron or two-charge system using the charge separation and signed energy termsexplain how electromagnetic energy emission changes total energy and therefore the orbital radius in an attractive two-charge modelinterpret orbital energy evidence to connect changes in potential energy, orbital speed and associated de Broglie wavelength
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≫