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### Part (a): Speed of the Proton
1. Known Values:
- Mass of a proton ([tex]\( m_p \)[/tex]): [tex]\( 1.67 \times 10^{-27} \)[/tex] kg
- Elementary charge ([tex]\( e \)[/tex]): [tex]\( 1.60 \times 10^{-19} \)[/tex] C
- Potential difference ([tex]\( V \)[/tex]): 124 V
2. Energy Considerations:
- When a proton is accelerated through a potential difference of 124 volts, it gains kinetic energy equal to the electrical potential energy provided by the voltage.
- The gained kinetic energy ([tex]\( KE \)[/tex]) can be expressed as [tex]\( KE = eV \)[/tex].
3. Kinetic Energy Formula:
- [tex]\( KE = \frac{1}{2} m_p v_p^2 \)[/tex]
- Equate the two expressions for kinetic energy: [tex]\( eV = \frac{1}{2} m_p v_p^2 \)[/tex].
4. Solve for Speed ([tex]\( v_p \)[/tex]):
- Rearrange to solve for [tex]\( v_p \)[/tex]:
[tex]\[ v_p = \sqrt{\frac{2 e V}{m_p}} \][/tex]
5. Substitute Known Values:
- [tex]\( e = 1.60 \times 10^{-19} \)[/tex] C
- [tex]\( V = 124 \)[/tex] V
- [tex]\( m_p = 1.67 \times 10^{-27} \)[/tex] kg
6. Calculate:
- Plugging in the known values, the speed of the proton is approximately:
[tex]\[ v_p \approx 154144.34 \, \text{m/s} \][/tex]
### Part (b): Speed of the Electron
1. Known Values:
- Mass of an electron ([tex]\( m_e \)[/tex]): [tex]\( 9.11 \times 10^{-31} \)[/tex] kg
- Elementary charge ([tex]\( e \)[/tex]): [tex]\( 1.60 \times 10^{-19} \)[/tex] C
- Potential difference ([tex]\( V \)[/tex]): 124 V
2. Energy Considerations:
- When an electron is accelerated through a potential difference of 124 volts, it gains kinetic energy equal to the electrical potential energy provided by the voltage.
- The gained kinetic energy ([tex]\( KE \)[/tex]) can be expressed as [tex]\( KE = eV \)[/tex].
3. Kinetic Energy Formula:
- [tex]\( KE = \frac{1}{2} m_e v_e^2 \)[/tex]
- Equate the two expressions for kinetic energy: [tex]\( eV = \frac{1}{2} m_e v_e^2 \)[/tex].
4. Solve for Speed ([tex]\( v_e \)[/tex]):
- Rearrange to solve for [tex]\( v_e \)[/tex]:
[tex]\[ v_e = \sqrt{\frac{2 e V}{m_e}} \][/tex]
5. Substitute Known Values:
- [tex]\( e = 1.60 \times 10^{-19} \)[/tex] C
- [tex]\( V = 124 \)[/tex] V
- [tex]\( m_e = 9.11 \times 10^{-31} \)[/tex] kg
6. Calculate:
- Plugging in the known values, the speed of the electron is approximately:
[tex]\[ v_e \approx 6599737.21 \, \text{m/s} \][/tex]
### Summary
- (a) The speed of the proton is approximately [tex]\( 154144.34 \)[/tex] m/s.
- (b) The speed of the electron is approximately [tex]\( 6599737.21 \)[/tex] m/s.
### Part (a): Speed of the Proton
1. Known Values:
- Mass of a proton ([tex]\( m_p \)[/tex]): [tex]\( 1.67 \times 10^{-27} \)[/tex] kg
- Elementary charge ([tex]\( e \)[/tex]): [tex]\( 1.60 \times 10^{-19} \)[/tex] C
- Potential difference ([tex]\( V \)[/tex]): 124 V
2. Energy Considerations:
- When a proton is accelerated through a potential difference of 124 volts, it gains kinetic energy equal to the electrical potential energy provided by the voltage.
- The gained kinetic energy ([tex]\( KE \)[/tex]) can be expressed as [tex]\( KE = eV \)[/tex].
3. Kinetic Energy Formula:
- [tex]\( KE = \frac{1}{2} m_p v_p^2 \)[/tex]
- Equate the two expressions for kinetic energy: [tex]\( eV = \frac{1}{2} m_p v_p^2 \)[/tex].
4. Solve for Speed ([tex]\( v_p \)[/tex]):
- Rearrange to solve for [tex]\( v_p \)[/tex]:
[tex]\[ v_p = \sqrt{\frac{2 e V}{m_p}} \][/tex]
5. Substitute Known Values:
- [tex]\( e = 1.60 \times 10^{-19} \)[/tex] C
- [tex]\( V = 124 \)[/tex] V
- [tex]\( m_p = 1.67 \times 10^{-27} \)[/tex] kg
6. Calculate:
- Plugging in the known values, the speed of the proton is approximately:
[tex]\[ v_p \approx 154144.34 \, \text{m/s} \][/tex]
### Part (b): Speed of the Electron
1. Known Values:
- Mass of an electron ([tex]\( m_e \)[/tex]): [tex]\( 9.11 \times 10^{-31} \)[/tex] kg
- Elementary charge ([tex]\( e \)[/tex]): [tex]\( 1.60 \times 10^{-19} \)[/tex] C
- Potential difference ([tex]\( V \)[/tex]): 124 V
2. Energy Considerations:
- When an electron is accelerated through a potential difference of 124 volts, it gains kinetic energy equal to the electrical potential energy provided by the voltage.
- The gained kinetic energy ([tex]\( KE \)[/tex]) can be expressed as [tex]\( KE = eV \)[/tex].
3. Kinetic Energy Formula:
- [tex]\( KE = \frac{1}{2} m_e v_e^2 \)[/tex]
- Equate the two expressions for kinetic energy: [tex]\( eV = \frac{1}{2} m_e v_e^2 \)[/tex].
4. Solve for Speed ([tex]\( v_e \)[/tex]):
- Rearrange to solve for [tex]\( v_e \)[/tex]:
[tex]\[ v_e = \sqrt{\frac{2 e V}{m_e}} \][/tex]
5. Substitute Known Values:
- [tex]\( e = 1.60 \times 10^{-19} \)[/tex] C
- [tex]\( V = 124 \)[/tex] V
- [tex]\( m_e = 9.11 \times 10^{-31} \)[/tex] kg
6. Calculate:
- Plugging in the known values, the speed of the electron is approximately:
[tex]\[ v_e \approx 6599737.21 \, \text{m/s} \][/tex]
### Summary
- (a) The speed of the proton is approximately [tex]\( 154144.34 \)[/tex] m/s.
- (b) The speed of the electron is approximately [tex]\( 6599737.21 \)[/tex] m/s.
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