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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

When a radiation of wavelength \(3000 \AA\) is incident on metal, $1.85 \mathrm{~V}$ stopping potential is obtained. What will be threshed wave length of metal? $\left\\{\mathrm{h}=66 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(4539 \AA\) (B) \(3954 \AA\) (C) \(5439 \AA\) (D) \(4395 \AA\)

Power produced by a star is \(4 \times 10^{28} \mathrm{~W}\). If the average wavelength of the emitted radiations is considered to be \(4500 \AA\) the number of photons emitted in \(1 \mathrm{~s}\) is \(\ldots \ldots\) (A) \(1 \times 10^{45}\) (B) \(9 \times 10^{46}\) (C) \(8 \times 10^{45}\) (D) \(12 \times 10^{46}\)

Calculate the energy of a photon of radian wavelength \(6000 \AA\) in \(\mathrm{eV}\) (A) \(20.6 \mathrm{eV}\) (B) \(2.06 \mathrm{eV}\) (C) \(1.03 \mathrm{eV}\) (D) \(4.12 \mathrm{eV}\)

A body of mass \(200 \mathrm{~g}\) moves at the speed of $5 \mathrm{~m} / \mathrm{hr}$. So deBroglie wavelength related to it is of the order........ \(\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\) (A) \(10^{-10} \mathrm{~m}\) (B) \(10^{-20} \mathrm{~m}\) (C) \(10^{-30} \mathrm{~m}\) (D) \(10^{-40} \mathrm{~m}\)

If ratio of threshold frequencies of two metals is \(1: 3\), ratio of their work functions is \(\ldots \ldots\) (A) \(1: 3\) (B) \(3: 1\) (C) \(4: 16\) (D) \(16: 4\)
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