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Chapter 17: Problem 2358

Photoelectric effect is obtained on metal surface for a light having frequencies \(\mathrm{f}_{1} \& \mathrm{f}_{2}\) where \(\mathrm{f}_{1}>\mathrm{f}_{2}\). If ratio of maximum kinetic energy of emitted photo electrons is \(1: \mathrm{K}\), so threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) $\left\\{\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (B) $\left\\{\left(\mathrm{Kf}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (C) $\left\\{\left(\mathrm{K} \mathrm{f}_{2}-\mathrm{f}_{1}\right) /(\mathrm{K}-1)\right\\}$ (D) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)

Short Answer

Expert verified
The threshold frequency for the metal surface is \(f_0 = \left\\{\left(Kf_1 - f_2\right) / (K - 1)\right\\}\).

Step by step solution

01

Write down the given information

We are given the frequencies \(f_1\), \(f_2\) and the ratio of maximum kinetic energies \(1:K\). Let the maximum kinetic energy at frequency \(f_1\) be \(E_1\) and at frequency \(f_2\) be \(E_2\). Based on the given information: \[ \frac{E_1}{E_2} = \frac{1}{K} \]
02

Use the photoelectric effect formula

According to the photoelectric effect formula: For frequency \(f_1\), \[ E_1 = h(f_1 - f_0) \] For frequency \(f_2\), \[ E_2 = h(f_2 - f_0) \]
03

Plug energy ratio into the equations

We can plug the energy ratio into the equations: \[ \frac{E_1}{E_2} = \frac{h(f_1 - f_0)}{h(f_2 - f_0)} = \frac{1}{K} \]
04

Solve for the threshold frequency

Now we will solve the equation for the threshold frequency \(f_0\): \[ \frac{f_1 - f_0}{f_2 - f_0} = \frac{1}{K} \] Multiply both sides by \(K(f_2 - f_0)\): \[ K(f_1 - f_0) = f_2 - f_0 \] Now, solve for \(f_0\): \[ f_0 = \frac{Kf_1 - f_2}{K - 1} \] So the threshold frequency for the metal surface is: \[ f_0 = \left\\{\left(Kf_1 - f_2\right) / (K - 1)\right\\} \] The correct answer is (B).

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Most popular questions from this chapter

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: Photo-electric effect can take place only when frequency is greater than threshold frequency \(\left(f_{0}\right)\) Reason: Electron is Fermions and photon is boson. (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) d

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (I) Energy of photon of wavelength \(\lambda\) is (P) \((\mathrm{E} / \mathrm{p})\) (II) The de Broglie wavelength associated (Q) \(\left(\mathrm{hf} / \mathrm{c}^{2}\right)\) with particle of momentum \(\mathrm{P}\) is (II) Mass of photon in motion is (R) (hc \(/ \lambda\) ) (IV) The velocity of photon of energy (S) \((\mathrm{h} / \mathrm{p})\) \(\mathrm{E}\) and momentum \(\mathrm{P}\) is (A) I - P, II - Q. III - R, IV - S (B) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$ (C) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{P}_{3} \mathrm{IV}-\mathrm{Q}$ (D) $\mathrm{I}-\mathrm{S}, \mathrm{II}-\mathrm{R}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$

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) \(2000 \mathrm{eV}\)

An electron moving with velocity \(0.6 \mathrm{c}\), then de-brogly wavelength associated with is \(\ldots \ldots \ldots\) (rest mars of electron, \(\mathrm{m}_{0}=9.1 \times 10^{-31}(\mathrm{k} / \mathrm{s})\) \(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) (A) \(3.24 \times 10^{-12} \mathrm{~m}\) (B) \(32.4 \times 10^{-12} \mathrm{~m}\) (C) \(320 \times 10^{-12} \mathrm{~m}\) (D) \(3.29 \times 10^{-14} \mathrm{~m}\)

The uncertainty in position of a particle is same as its de-Broglie wavelength, uncertainty in its momentum is \(\ldots \ldots\) (A) \((\overline{\mathrm{h}} / \lambda)\) (B) \((2 \overline{\mathrm{h}} / 3 \lambda)\) (C) \((\lambda \lambda) \overline{\mathrm{h}}\) (D) \((3 N 2 \lambda) \overline{\mathrm{h}}\)
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