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Chapter 17: Problem 2359

Frequency of incident light on body is \(\mathrm{f}\). Threshold frequency of body is \(f_{0}\). Maximum velocity of electron \(=\ldots \ldots \ldots\).. where \(m\) is mass of electron. (A) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]^{(1 / 2)}$ (B) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]$ (C) \([2 \mathrm{hf} / \mathrm{m}]^{(1 / 2)}\) (D) \(\mathrm{h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\)

Short Answer

Expert verified
The maximum velocity of the electron is given by the formula: \(v = \sqrt{\frac{2h(f - f_{0})}{m}}\). Therefore, the correct answer is (A).

Step by step solution

01

Recall Einstein's Photoelectric Equation

According to Einstein's photoelectric equation, the maximum kinetic energy (K.E.) of the emitted electron is given by: $$K.E. = h(f - f_{0})$$ Here, h is Planck's constant, f is the frequency of the incident light, and \(f_{0}\) is the threshold frequency.
02

Express Kinetic Energy in terms of Mass and Velocity

We can express the kinetic energy in terms of mass (m) and velocity (v) as follows: $$K.E. = \frac{1}{2}mv^2$$
03

Equate the two expressions of Kinetic Energy

Now, we will equate the two expressions for kinetic energy and solve for the maximum velocity (v): $$\frac{1}{2}mv^2 = h(f - f_{0})$$
04

Solve for the Maximum Velocity

Isolate the variable v by multiplying both sides of the equation by the reciprocal of \(\frac{1}{2}m\): $$v^2 = \frac{2h(f - f_{0})}{m}$$ Now, take the square root of both sides to find the maximum velocity: $$v = \sqrt{\frac{2h(f - f_{0})}{m}}$$
05

Find the correct answer from the multiple-choice options

Comparing the derived equation to the given options: (A) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]^{(1 / 2)}$ (B) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]$ (C) \([2 \mathrm{hf} / \mathrm{m}]^{(1 / 2)}\) (D) \(\mathrm{h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\) We can see that our derived equation is the same as option (A): $$v = \sqrt{\frac{2h(f - f_{0})}{m}}$$ So, the correct answer is (A).

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Most popular questions from this chapter

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) $\left\\{(9.46 \mathrm{~h}) /\left(\mathrm{m}_{\circ} \mathrm{C}\right)\right\\}$

A proton and an \(\propto\) -particle are passed through same potential difference. If their initial velocity is zero, the ratio of their de Broglie's wavelength after getting accelerated is \(\ldots \ldots\) (A) \(1: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(2 \sqrt{2}: 1\)

An electron is accelerated under a potential difference of \(64 \mathrm{~V}\), the de Broglie wave length associated with electron is $=\ldots \ldots \ldots \ldots . \AA$ (Use charge of election $=1.6 \times 10^{-19} \mathrm{C},=9.1 \times 10^{-31} \mathrm{Kg}$, mass of electrum \(\mathrm{h}=6.6623 \times 10^{-43} \mathrm{~J}\).sec \()\) (A) \(4.54\) (B) \(3.53\) (C) \(2.53\) (D) \(1.534\)

Direction Read the following question choose if: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) Assertion is true but the Reason is false. (d) Both Assertion and Reason is false. Assertion: Metals like Na or \(\mathrm{K}\), emit electrons even when visible lights fall on them. Reason: This is because their work function is low. (A) a (B) \(\mathrm{b}\) (C) \(\mathrm{c}\) (D) d

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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