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Chapter 17: Problem 2389

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) \(16: 4\)

Short Answer

Expert verified
The ratio of the work functions of the two metals is \(1:3\), which corresponds to the answer (A).

Step by step solution

01

Write the photoelectric effect equation for both metals

The photoelectric effect equation is given by: \(E = h \nu = \phi + KE\) where: - \(E\) is the energy of the incident photon - \(h\) is the Planck's constant - \(\nu\) is the frequency of the incident photon - \(\phi\) is the work function of the metal - \(KE\) is the kinetic energy of the ejected electron Now let's write the photoelectric effect equation for both the metals: For metal 1: \(E_1 = h \nu_1 = \phi_1 + KE_1\) For metal 2: \(E_2 = h \nu_2 = \phi_2 + KE_2\)
02

Use the given ratio of their threshold frequencies

The given ratio of threshold frequencies of the two metals is 1:3. Therefore, we can write it as: \(\frac{\nu_1}{\nu_2} = \frac{1}{3}\) Now let's substitute this ratio in the photoelectric effect equations to find the work functions: \(E_1 = h \cdot \frac{1}{3} \nu_2 = \phi_1 + KE_1\) \(E_2 = h \nu_2 = \phi_2 + KE_2\)
03

Determine the ratio of their work functions

Now we need to find the ratio of their work functions, which can be written as: \(\frac{\phi_1}{\phi_2}\) Since \(E_1 = \frac{1}{3} E_2\), we can write the equations as: \(\phi_1 = \frac{1}{3} \phi_2\) \(\frac{\phi_1}{\phi_2} = \frac{\frac{1}{3} \phi_2}{\phi_2}\) Simplifying the above equation, we get: \(\frac{\phi_1}{\phi_2} = \frac{1}{3}\)
04

Identify the correct answer

Now we can see that the ratio of the work functions of the two metals is \(1:3\), which corresponds to the answer (A).

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Most popular questions from this chapter

If alpha particle and deuteron move with velocity \(\mathrm{V}\) and $2 \mathrm{v}$, the ratio of their de-Brogle wavelength will be............ deuteron (A) \(1: \sqrt{2}\) (B) \(2: 1\) (C) \(1: 1\) (D) \(\sqrt{2}: 1\)

Photoelectric effect on surface is found for frequencies $5.5 \times 10^{8} \mathrm{MHz}\( and \)4.5 \times 10^{8} \mathrm{MHz}$ If ratio of maximum kinetic energies of emitted photo electrons is \(1: 5\), threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) \(7.55 \times 10^{8} \mathrm{MHz}\) (B) \(4.57 \times 10^{8} \mathrm{MHz}\) (C) \(9.35 \times 10^{8} \mathrm{MHz}\) (D) \(5.75 \times 10^{8} \mathrm{MHz}\)

\(2 \mathrm{nW}\) light of wave length \(4400 \AA\) is incident on photo sensitive surface of \(\mathrm{Cs}\). If quantum efficiency is \(0.5 \%\), what will be the value of photoelectric current? (A) \(1.56 \times 10^{-6} \mu \mathrm{A}\) (B) \(2.56 \times 10^{-6} \mu \mathrm{A}\) (C) \(4.56 \times 10^{-6} \mu \mathrm{A}\) (D) \(3.56 \times 10^{-6} \mu \mathrm{A}\)

Work function of metal is \(2 \mathrm{eV}\). Light of intensity $10^{-5} \mathrm{Wm}^{-2}\( is incident on \)2 \mathrm{~cm}^{2}\( area of it. If \)10^{17}$ electrons of these metals absorb the light, in how much time does the photo electric effectc start? Consider the waveform of incident light. (A) \(1.4 \times 10^{7} \mathrm{sec}\) (B) \(1.5 \times 10^{7} \mathrm{sec}\) (C) \(1.6 \times 10^{7} \mathrm{sec}\) (D) \(1.7 \times 10^{7} \mathrm{sec}\)

Particle \(\mathrm{A}\) and \(\mathrm{B}\) have electric charge \(+\mathrm{q}\) and \(+4 \mathrm{q} .\) Both have mass \(\mathrm{m}\). If both are allowed to fall under the same p.d., ratio of velocities $\left(\mathrm{V}_{\mathrm{A}} / \mathrm{V}_{\mathrm{B}}\right)=\ldots \ldots \ldots \ldots \ldots$ (A) \(2: 1\) (B) \(1: 2\) (C) \(1: 4\) (D) \(4: 1\)
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