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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

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) \(0.75 \mathrm{~nm}\)

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) \(24.8\)

Read the paragraph carefully and select the proper choice from given multiple choices. According to Einstein when a photon of light of frequency for wavelength \(\lambda\) is incident on a photo sensitive metal surface of work function \(\Phi\). Where \(\Phi<\mathrm{hf}\) (here \(\mathrm{h}\) is Plank's constant) then the emission of photo-electrons place takes place. The maximum K.E. of emitted photo electrons is given by $\mathrm{K}_{\max }=\mathrm{hf}-\Phi .\( If the there hold frequency of metal is \)\mathrm{f}_{0}$ then \(\mathrm{hf}_{0}=\Phi\) (i) A metal of work function \(3.3 \mathrm{eV}\) is illuminated by light of wave length \(300 \mathrm{~nm}\). The maximum \(\mathrm{K} . \mathrm{E}>\) of photo- electrons is $=\ldots \ldots \ldots . \mathrm{eV} .\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}\right)$ (A) \(0.825\) (B) \(0.413\) (C) \(1.32\) (D) \(1.65\) (ii) Stopping potential of emitted photo-electron is $=\ldots \ldots . \mathrm{V}$. (A) \(0.413\) (B) \(0.825\) (C) \(1.32\) (D) \(1.65\) (iii) The threshold frequency fo $=\ldots \ldots \ldots \times 10^{14} \mathrm{~Hz}$. (A) \(4.0\) (B) \(4.2\) (C) \(8.0\) (D) \(8.4\)

With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is \(0.4 \AA\) (A) \(9410 \mathrm{~V}\) (B) \(94.10 \mathrm{~V}\) (C) \(9.140 \mathrm{~V}\) (D) \(941.0 \mathrm{~V}\)

de-Broglie wavelength of atom at T \(\mathrm{K}\) absolute temperature will be \(\ldots \ldots \ldots\) (A) \([\mathrm{h} /\\{\mathrm{mkT}\\}]\) (B) \([\mathrm{h} /\\{\sqrt{3} \mathrm{mKT}\\}]\) (C) \([\\{\sqrt{2} \mathrm{mKT}\\} / \mathrm{h}]\) (D) \(\sqrt{(2 \mathrm{mKT})}\)
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