Category: HC Verma Part 2: Modern Physics
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HC Verma Chapter 41 Problem 10 — Paschen Minimum Breakdown Voltage
Problem Statement Explain qualitatively why the breakdown voltage in a gas discharge tube has a minimum (Paschen’s law) as a function of $pd$ (pressure × gap distance). Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the…
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HC Verma Chapter 41 Problem 9 — Number Density and Mean Free Path
Problem Statement At pressure $p = 1\times10^{-3}$ mmHg and temperature $T = 300$ K, find the number density of gas molecules. (1 mmHg $= 133.3$ Pa) Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant…
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HC Verma Chapter 41 Problem 8 — Minimum Potential Difference for Ionization
Problem Statement The ionization energy of helium is 24.6 eV. Find the minimum potential difference through which an electron must be accelerated to ionize helium atoms by collision. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying…
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HC Verma Chapter 41 Problem 7 — Deflection of Electron in Electric Field
Problem Statement Determine the electric field for the configuration described: Electrons with speed $v = 3\times10^7$ m/s enter a uniform electric field $E = 10^4$ V/m perpendicular to their velocity. Find the transverse deflection after travelling $L = 0.1$ m horizontally. Time to cross: $t = L/v$; transverse acceleration: $a = eE/m_e$; deflection: $y =…
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HC Verma Chapter 41 Problem 6 — Electron Velocity from Potential Difference
Problem Statement Solve the kinematics problem: An electron is accelerated from rest through a potential difference of 200 V. Calculate its final kinetic energy and speed. $KE = eV$; $v = \sqrt{2KE/m_e}$ Step 1: $KE = eV = 1.6\times10^{-19}\times200 = 3.2\times10^{-17}$ J $= 200$ eV Step 2: $$v = \sqrt{\frac{2\times3.2\times10^{-17}}{9.1\times10^ Given Information See problem statement…
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HC Verma Chapter 41 Problem 5 — Townsend Discharge and Secondary Emission
Problem Statement In a Townsend discharge, the first Townsend ionization coefficient is $\alpha = 10$ cm$^{-1}$ and the distance between electrodes is $d = 1$ cm. If the number of electrons leaving the cathode per second is $n_0 = 10^6$, find the number of electrons reaching the anode. Given Information See problem statement for all…
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HC Verma Chapter 41 Problem 4 — e/m Ratio of Electron
Problem Statement In a Thomson experiment, electrons pass undeflected when $E = 3\times10^4$ V/m and $B = 1.5\times10^{-3}$ T are applied together. When only $B$ is applied, the radius of circular path is $r = 0.12$ m. Find $e/m$ for the electron. Given Information See problem statement for all given quantities. Physical Concepts & Formulas…
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HC Verma Chapter 41 Problem 3 — Cathode Ray Deflection in Electric Field
Problem Statement Determine the electric field for the configuration described: Electrons are accelerated through a potential difference of $V = 1000$ V. Find their speed after acceleration. Energy conservation: $eV = \frac{1}{2}m_e v^2 \Rightarrow v = \sqrt{2eV/m_e}$ $m_e = 9.1\times10^{-31}$ kg, $e = 1.6\times10^{-19}$ C Step 1: $$v = \sqrt{\frac{2eV}{m_e}} = \sqrt{\frac{2\ti Given Information See…
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HC Verma Chapter 41 Problem 2 — Mean Free Path of Electrons
Problem Statement The mean free path of electrons in a gas discharge tube is $\lambda = 0.5$ mm. If the electrons need energy of 12 eV to ionize the gas, find the minimum electric field required. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles…
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HC Verma Chapter 41 Problem 1 — Ionization Energy of Hydrogen
Problem Statement Solve the work-energy problem: Find the ionization energy of hydrogen atom, given the ground state energy is $-13.6$ eV. Ionization energy = energy needed to remove electron from ground state to $n=\infty$ Step 1: At $n=\infty$, energy = 0. Ground state energy = $-13.6$ eV. Step 2: Energy needed: $$E_{ion} = 0 –…