| $E = 1\ \text{MeV}$ | | $E = 1\ \text{MeV}$ |
| | | |
| Rest masses | | Rest masses |
| | | |
| $m_e c^2 = 0.511\ \text{MeV}, m_p c^2 = 938\ \text{MeV}$ | | $m_e c^2 = 0.511\ \text{MeV}, m_p c^2 = 938\ \text{MeV}$ |
| | | |
| Relationship between deflection and momentum | | Relationship between deflection and momentum |
| | | |
| In a plane capacitor with a uniform transverse electric field, a particle of charge e receives a transverse impulse | | In a plane capacitor with a uniform transverse electric field, a particle of charge e receives a transverse impulse |
| $\Delta p_\perp = eE_\text{field}\,\Delta t$ The flight time is $\Delta t \approx L / v $ | | $\Delta p_\perp = eE_\text{field}\,\Delta t$ The flight time is $\Delta t \approx L / v $ |
| | | |
| where L is the length of the plates and v is the initial velocity. For small angles, the deflection is: | | where L is the length of the plates and v is the initial velocity. For small angles, the deflection is: |
| | | |
| $\alpha \approx \frac{\Delta p_\perp}{p} \propto \frac{1}{p v}$ | | $\alpha \approx \frac{\Delta p_\perp}{p} \propto \frac{1}{p v}$ |
| | | |
| where p is the initial momentum and v is the velocity. The same geometry and field imply$ \alpha \propto 1/(p v)$ | | where p is the initial momentum and v is the velocity. The same geometry and field imply$ \alpha \propto 1/(p v)$ |
| | | |
| Calculation of p v for a particle with kinetic energy E | | Calculation of p v for a particle with kinetic energy E |
| | | |
| In relativity, the total energy is $E_{\text{total}} = E + mc^2$ The momentum satisfies$ (pc)^2 = E_{\text{total}}^2 - (mc^2)^2$ The velocity is$ v/c = pc / E_{\text{total}} $ | | In relativity, the total energy is $E_{\text{total}} = E + mc^2$ The momentum satisfies$ (pc)^2 = E_{\text{total}}^2 - (mc^2)^2$ The velocity is$ v/c = pc / E_{\text{total}} $ |
| | | |
| Multiplying: | | Multiplying: |
| | | |
| $p v = \frac{p^2 c^2}{E_{\text{total}}} = \frac{(E + mc^2)^2 - (mc^2)^2}{E + mc^2} = \frac{E(E + 2mc^2)}{E + mc^2}$ | | $p v = \frac{p^2 c^2}{E_{\text{total}}} = \frac{(E + mc^2)^2 - (mc^2)^2}{E + mc^2} = \frac{E(E + 2mc^2)}{E + mc^2}$ |
| | | |
| Thus, the product pv is expressed in terms of E and the mass m: | | Thus, the product pv is expressed in terms of E and the mass m: |
| | | |
| $p v = \frac{E(E + 2mc^2)}{E + mc^2}$ | | $p v = \frac{E(E + 2mc^2)}{E + mc^2}$ |
| | | |
| Comparison of the angles | | Comparison of the angles |
| | | |
| Since$ \alpha \propto 1/(p v)$ the ratio between the deflection angle of the electron and that of the proton is the inverse of the ratio of their p v products: | | Since$ \alpha \propto 1/(p v)$ the ratio between the deflection angle of the electron and that of the proton is the inverse of the ratio of their p v products: |
| | | |
| $\frac{\alpha_e}{\alpha_p} = \frac{(p v)_p}{(p v)_e}$ | | $\frac{\alpha_e}{\alpha_p} = \frac{(p v)_p}{(p v)_e}$ |
| | | |
| Substituting the expressions: | | Substituting the expressions: |
| | | |
| $\frac{\alpha_e}{\alpha_p} = \frac{ \dfrac{E(E + 2m_pc^2)}{E + m_pc^2} }{ \dfrac{E(E + 2m_ec^2)}{E + m_ec^2} } | | $\frac{\alpha_e}{\alpha_p} = \frac{ \dfrac{E(E + 2m_pc^2)}{E + m_pc^2} }{ \dfrac{E(E + 2m_ec^2)}{E + m_ec^2} } |
| = \frac{E + m_ec^2}{E + 2m_ec^2} \cdot \frac{E + 2m_pc^2}{E + m_pc^2}$ | | = \frac{E + m_ec^2}{E + 2m_ec^2} \cdot \frac{E + 2m_pc^2}{E + m_pc^2}$ |
| | | |
| and for small angles $(\alpha_p \ll 1, \alpha_e \ll 1)$ we can approximate$ \tan \alpha \approx \alpha$ | | and for small angles $(\alpha_p \ll 1, \alpha_e \ll 1)$ we can approximate$ \tan \alpha \approx \alpha$ |
| | | |
| Numerical calculation | | Numerical calculation |
| | | |
| With$ E = 1\ \text{MeV}, m_ec^2 = 0.511\ \text{MeV}, m_pc^2 = 938\ \text{MeV}$ | | With$ E = 1\ \text{MeV}, m_ec^2 = 0.511\ \text{MeV}, m_pc^2 = 938\ \text{MeV}$ |
| | | |
| $\frac{m_ec^2 + E}{2m_ec^2 + E} = \frac{0.511 + 1}{1.022 + 1} = \frac{1.511}{2.022} \approx 0.7473$ | | $\frac{m_ec^2 + E}{2m_ec^2 + E} = \frac{0.511 + 1}{1.022 + 1} = \frac{1.511}{2.022} \approx 0.7473$ |
| | | |
| $\frac{2m_pc^2 + E}{m_pc^2 + E} = \frac{1876 + 1}{938 + 1} = \frac{1877}{939} \approx 1.9989 \approx 2.00$. | | $\frac{2m_pc^2 + E}{m_pc^2 + E} = \frac{1876 + 1}{938 + 1} = \frac{1877}{939} \approx 1.9989 \approx 2.00$. |
| | | |
| The product is$ 0.7473 \times 1.9989 \approx 1.494$. | | The product is$ 0.7473 \times 1.9989 \approx 1.494$. |
| | | |
| For $\alpha_p = 0.05\ \text{rad}$: | | For $\alpha_p = 0.05\ \text{rad}$: |
| | | |
| $\alpha_e \approx 1.494 \times 0.05 \approx 0.0747\ \text{rad} \approx 0.075\ \text{rad}$. | | $\alpha_e \approx 1.494 \times 0.05 \approx 0.0747\ \text{rad} \approx 0.075\ \text{rad}$. |
| | | |
| #### Answer | | #### Answer |
| | | |
| $\boxed{\tan \alpha_e = \frac{m_ec^2 + E}{2m_ec^2 + E} \cdot \frac{2m_pc^2 + E}{m_pc^2 + E} \cdot \tan \alpha_p}$ | | $\boxed{\tan \alpha_e = \frac{m_ec^2 + E}{2m_ec^2 + E} \cdot \frac{2m_pc^2 + E}{m_pc^2 + E} \cdot \tan \alpha_p}$ |
| | | |
| $\boxed{\alpha_e \approx 0.075\ \text{rad}}$. | | $\boxed{\alpha_e \approx 0.075\ \text{rad}}$. |
| | | |
| The electron deflects approximately 50% more than the proton because, at the same kinetic energy, its smaller mass makes the product p v smaller, thereby increasing the deflection | | The electron deflects approximately 50% more than the proton because, at the same kinetic energy, its smaller mass makes the product p v smaller, thereby increasing the deflection |
