An electron and a proton starting from rest get accelerated through potential difference of 100 kV. The final speeds of the electron and the proton are V e and V p respectively. Which one of the following relations is correct?
V e> V p
When a charged particle is accelerated by a potential difference, the work done by the electric field on the particle is converted into kinetic energy. This principle helps us determine the final speed of the particle.
Let's consider a charged particle with charge magnitude \(|q|\) and mass \(m\). When it is accelerated through a potential difference \(V\), the work done (\(W\)) on the particle is given by:
\[W = |q|V\]
If the particle starts from rest, its initial kinetic energy is zero. The final kinetic energy (\(KE_f\)) after being accelerated through the potential difference is equal to the work done:
\[KE_f = W\]
The kinetic energy of a particle with speed \(v\) is given by:
\[KE = \frac{1}{2}mv^2\]
So, equating the work done and the final kinetic energy:
\[|q|V = \frac{1}{2}mv^2\]
We can rearrange this equation to find the final speed \(v\):
\[v^2 = \frac{2|q|V}{m}\]
\[v = \sqrt{\frac{2|q|V}{m}}\]
We are given that an electron and a proton are accelerated through the same potential difference, \(V = 100 \text{ kV}\). Both particles start from rest.
Let's consider the electron:
The speed of the electron will be:
\[v_e = \sqrt{\frac{2e V}{m_e}}\]
Now let's consider the proton:
The speed of the proton will be:
\[v_p = \sqrt{\frac{2e V}{m_p}}\]
We have the expressions for \(v_e\) and \(v_p\):
\[v_e = \sqrt{\frac{2e V}{m_e}} \quad \text{and} \quad v_p = \sqrt{\frac{2e V}{m_p}}\]
Notice that the term \(2eV\) is the same for both the electron and the proton, as they have the same magnitude of charge (\(e\)) and are accelerated through the same potential difference (\(V\)).
The speed is inversely proportional to the square root of the mass:
\[v \propto \frac{1}{\sqrt{m}}\]
This means that if the mass is smaller, the speed will be larger, assuming all other factors are constant.
We know that the mass of an electron (\(m_e\)) is significantly smaller than the mass of a proton (\(m_p\)). Approximately, \(m_p \approx 1836 \times m_e\).
Since \(m_e < m_p\), the denominator in the expression for \(v_e\) is smaller than the denominator in the expression for \(v_p\). Therefore, the value under the square root is larger for the electron.
\[\frac{1}{m_e} > \frac{1}{m_p}\]
\[\frac{2eV}{m_e} > \frac{2eV}{m_p}\]
\[\sqrt{\frac{2eV}{m_e}} > \sqrt{\frac{2eV}{m_p}}\]
Thus, the speed of the electron is greater than the speed of the proton:
\[v_e > v_p\]
This relation matches one of the given options.
| Particle | Charge Magnitude | Mass | Potential Difference | Speed Formula |
|---|---|---|---|---|
| Electron | \(e\) | \(m_e\) | \(V\) | \(v_e = \sqrt{\frac{2eV}{m_e}}\) |
| Proton | \(e\) | \(m_p\) | \(V\) | \(v_p = \sqrt{\frac{2eV}{m_p}}\) |
Comparing \(v_e\) and \(v_p\), we see that since \(m_e < m_p\), it follows that \(v_e > v_p\).
| Concept | Formula / Principle | Application Here |
|---|---|---|
| Work done by Electric Field | \(W = qV\) (for magnitude) | Energy gained by particle |
| Kinetic Energy | \(KE = \frac{1}{2}mv^2\) | Energy of motion |
| Energy Conservation | \(|q|V = \frac{1}{2}mv^2\) | Potential energy converted to kinetic energy |
| Speed relation to mass | \(v \propto \frac{1}{\sqrt{m}}\) (for fixed energy) | Lighter particles are faster for same energy |
| Electron vs Proton Mass | \(m_e < m_p\) | Electron is much lighter than proton |
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