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Chapter 17: Problem 2408

The mass of a particle is 400 times than that of an electron and charge is double. The particle is accelerated by \(5 \mathrm{~V}\). Initially the particle remained at rest, then its final kinetic energy is \(\ldots \ldots \ldots\) (A) \(5 \mathrm{eV}\) (B) \(10 \mathrm{eV}\) (C) \(100 \mathrm{eV}\) (D) \(2000 \mathrm{eV}\)

Short Answer

Expert verified
The final kinetic energy of the particle after being accelerated is \(10 eV\).

Step by step solution

01

Calculate the work done by the electric field

Given: Mass of the electron, \(m_e = 9.1 \times 10^{-31} kg\) Charge of the electron, \(e = 1.6 \times 10^{-19} C\) Mass of the particle, \(m_p = 400 m_e\) Charge of the particle, \(q_p = 2e = 3.2 \times 10^{-19} C\) The particle is accelerated by a potential difference of 5 V. The work done by the electric field on the charged particle is given by: \[W = q_pV\]
02

Use the work-energy theorem to find the final kinetic energy of the particle

According to the work-energy theorem, the work done on a body is equal to the change in its kinetic energy. Since the initial velocity of the particle is 0, the change in kinetic energy is equal to the final kinetic energy, \(K_f\). \[K_f = W\] Now, substitute all the given values to find the final kinetic energy: \[K_f = q_pV = (3.2 \times 10^{-19} C)(5 V) = 16 \times 10^{-19} J\] Convert this value to electron volts (eV) using the conversion factor \(1 eV = 1.6 \times 10^{-19} J\): \[K_f = \frac{16 \times 10^{-19} J}{1.6 \times 10^{-19} J/eV} = 10 eV\] So, the final kinetic energy of the particle is \(10 eV\). Therefore, the correct option is (B).

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Most popular questions from this chapter

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (A) Planck's theory of quantum (p) Light energy \(=\mathrm{hv}\) (B) Einstein's theory of quanta (q) Angular momentum of electron in an orbit. (C) Bohr's stationary orbit (r) Oscillator energies (D) D-Broglie waves (s) Electron microscope (A) \((\mathrm{A}-\mathrm{p}),(\mathrm{b}-\mathrm{q}),(\mathrm{C}-\mathrm{r}),(\mathrm{D}-\mathrm{s})\) (B) \((\mathrm{A}-\mathrm{q}),(\mathrm{B}-\mathrm{r}),(\mathrm{C}-\mathrm{s}),(\mathrm{D}-\mathrm{p})\) (C) \((\mathrm{A}-\mathrm{r}),(\mathrm{B}-\mathrm{p}),(\mathrm{C}-\mathrm{q}),(\mathrm{D}-\mathrm{s})\) (D) \((\mathrm{A}-\mathrm{r}),(\mathrm{B}-\mathrm{p}),(\mathrm{C}-\mathrm{s}),(\mathrm{D}-\mathrm{q})\)

According to Rayleigh and Jeans the black body radiation in the cavity is system of (A) progressive electromagnetic waves (B) standing electromagnetic waves (C) electromagnetic waves of discrete (D) standing waves in lattice frequencies

Compare energy of a photon of X-rays having \(1 \AA\) wavelength with the energy of an electron having same de Broglie wavelength. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, 1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(8.24\) (B) \(2.48\) (C) \(82.4\) (D) \(24.8\)

De-Broglie wavelength of particle moving at a (1/4) th of speed of light having rest mass \(\mathrm{m}_{0}\) is \(\ldots \ldots \ldots\) (A) $\left\\{(3.87 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (B) $\left\\{(4.92 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (C) $\left\\{(7.57 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (D) $\left\\{(9.46 \mathrm{~h}) /\left(\mathrm{m}_{\circ} \mathrm{C}\right)\right\\}$

An electron is accelerated between two points having potential $20 \mathrm{~V}\( and \)40 \mathrm{~V}$, de-Broglic wavelength of electron is \(\ldots \ldots\) (A) \(0.75 \AA\) (B) \(7.5 \AA\) (C) \(2.75 \AA\) (D) \(0.75 \mathrm{~nm}\)
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