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Chapter 17: Problem 2396

Photons of energy \(1 \mathrm{eV}\) and \(2.5 \mathrm{ev}\) successively illuminate a metal, whose work function is \(0.5 \mathrm{eV}\), the ratio of maximum speed of emitted election is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(3: 1\) (D) \(1: 3\)

Short Answer

Expert verified
The ratio of maximum speeds of emitted electrons is \(1:2\). Therefore, the correct option is (A) \(1:2\).

Step by step solution

01

Write down the photoelectric effect formula

The photoelectric effect formula is given by: \(K.E = h\nu - W\) where \(K.E\) is the kinetic energy of the emitted electrons, \(h\) is the Planck's constant, \(\nu\) is the frequency of the incident photons, and \(W\) is the work function of the metal. We can rewrite the formula in terms of energy (in eV) as: \(K.E = E_\text{photon} - W\)
02

Calculate the kinetic energy for each case

We are given that the energy of the incident photons for the two cases are \(1 \mathrm{eV}\) and \(2.5 \mathrm{eV}\), and the work function of the metal is \(0.5 \mathrm{eV}\). Case 1: Photon energy \(1 \mathrm{eV}\) \(K.E_1 = E_{\text{photon 1}} - W = 1\mathrm{eV} - 0.5\mathrm{eV} = 0.5\mathrm{eV}\) Case 2: Photon energy \(2.5 \mathrm{eV}\) \(K.E_2 = E_{\text{photon 2}} - W = 2.5\mathrm{eV} - 0.5\mathrm{eV} = 2\mathrm{eV}\)
03

Use the kinetic energy formula to calculate maximum speed

We will now use the kinetic energy formula, which states that \(K.E = \frac{1}{2}mv^2\) where \(m\) is the mass of the electron and \(v\) is its speed. For the two cases, we get: Case 1: \(K.E_1 = \frac{1}{2}m(v_1)^2 \Rightarrow v_1 = \sqrt{\frac{2K.E_1}{m}}\) Case 2: \(K.E_2 = \frac{1}{2}m(v_2)^2 \Rightarrow v_2 = \sqrt{\frac{2K.E_2}{m}}\)
04

Calculate the ratio of maximum speeds

To calculate the required ratio, we divide the expressions for the maximum speeds in the two cases: \(\frac{v_1}{v_2} = \frac{\sqrt{\frac{2K.E_1}{m}}}{\sqrt{\frac{2K.E_2}{m}}} = \frac{\sqrt{K.E_1}}{\sqrt{K.E_2}}\) Using the values of \(K.E_1\) and \(K.E_2\) calculated in Step 2: \(\frac{v_1}{v_2} = \frac{\sqrt{0.5\mathrm{eV}}}{\sqrt{2\mathrm{eV}}} = \frac{\sqrt{1/2}}{\sqrt{2}} = \frac{1}{\sqrt{2^2}} = \frac{1}{2}\) This ratio can be written as \(1:2\).
05

Answer

The ratio of maximum speeds of emitted electrons is \(1:2\). Therefore, the correct option is (A) \(1:2\).

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Most popular questions from this chapter

If velocity of free electron is made double, change in its de-Broglie wavelength will be \(\ldots \ldots .\) (A) increase by \((\lambda / 2)\) (B) decrease by \((\lambda 2)\) (C) increase by \(2 \lambda\) (D) decrease \(2 \lambda\)

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) \(3.29 \times 10^{-14} \mathrm{~m}\)

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) \(\mathrm{h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\)

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) \(\sqrt{(2 \mathrm{mKT})}\)

At \(10^{\circ} \mathrm{C}\) temperature, de-Broglie wave length of atom is $0.4 \AA\(.If temperature of atom is increased by \)30^{\circ} \mathrm{C}$, what will be change in de-Broglie wavelength of atom? (A) decreases \(10^{-2} \AA\) (B) decreases \(2 \times 10^{-2} \AA\) (C) increases \(10^{-2} \AA\) (D) increases \(2 \times 10^{-2} \AA\)
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