Question:
An electron beam accelerated from rest through a potential difference of 5000 V in vacuum is allowed to impinge on a surface normally. The incident current is 50 \(\mu\)A and if the electron comes to rest on striking the surface, the force on it is
An electron beam accelerated from rest through a potential difference of 5000 V in vacuum is allowed to impinge on a surface normally. The incident current is 50 \(\mu\)A and if the electron comes to rest on striking the surface, the force on it is
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Force from an electron beam = momentum delivered per second = $n \cdot mv$, where $n$ is the number of electrons per second and $v$ is found from the accelerating potential.
Updated On: Apr 20, 2026
- \(1.1924 \times 10^{-8}\) N
- \(2.1 \times 10^{-8}\) N
- \(1.6 \times 10^{-8}\) N
- \(1.6 \times 10^{-6}\) N
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Force = rate of change of momentum = \(n \cdot mv\) per second, where \(n\) is number of electrons hitting per second.
Step 2: Detailed Explanation:
Speed of electrons: \(\frac{1}{2}mv^2 = eV \Rightarrow mv^2 = 2eV \Rightarrow mv = \sqrt{2meV}\). Number per second: \(n = I/e = 50\times10^{-6}/(1.6\times10^{-19}) = 3.125\times10^{14}\). Force \(= nmv = n\sqrt{2m_eV\cdot e}\): \[ = 3.125\times10^{14}\times\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times5000} = 1.1924\times10^{-8} \text{ N} \]
Step 3: Final Answer:
Force \(= 1.1924 \times 10^{-8}\) N.
Force = rate of change of momentum = \(n \cdot mv\) per second, where \(n\) is number of electrons hitting per second.
Step 2: Detailed Explanation:
Speed of electrons: \(\frac{1}{2}mv^2 = eV \Rightarrow mv^2 = 2eV \Rightarrow mv = \sqrt{2meV}\). Number per second: \(n = I/e = 50\times10^{-6}/(1.6\times10^{-19}) = 3.125\times10^{14}\). Force \(= nmv = n\sqrt{2m_eV\cdot e}\): \[ = 3.125\times10^{14}\times\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times5000} = 1.1924\times10^{-8} \text{ N} \]
Step 3: Final Answer:
Force \(= 1.1924 \times 10^{-8}\) N.
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