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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

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) \(5.75 \times 10^{8} \mathrm{MHz}\)

Work function of metal is \(2.5 \mathrm{eV}\) If wave length of light incident on metal plate is \(3000 \AA\), stopping potential of emitted electron will be....... $\left\\{\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(0.82 \mathrm{~V}\) (B) \(0.41 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(3.28 \mathrm{~V}\)

If electron is accelerated under \(50 \mathrm{KV}\) in microscope, find its de- Broglie wavelength. (A) \(5.485 \times 10^{-12} \mathrm{~m}\) (B) \(8.545 \times 10^{-12} \mathrm{~m}\) (C) \(4.585 \times 10^{-12} \mathrm{~m}\) (D) \(5.845 \times 10^{-12} \mathrm{~m}\)

With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is \(0.4 \AA\) (A) \(9410 \mathrm{~V}\) (B) \(94.10 \mathrm{~V}\) (C) \(9.140 \mathrm{~V}\) (D) \(941.0 \mathrm{~V}\)

Output power of He-Ne LASER of low energy is \(1.00 \mathrm{~mW}\). Wavelength of the light is \(632.8 \mathrm{~nm}\). What will be the number of photons emitted per second from this LASER? (A) \(8.31 \times 10^{15} \mathrm{~s}^{-1}\) (B) \(5.38 \times 10^{15} \mathrm{~s}^{-1}\) (C) \(1.83 \times 10^{15} \mathrm{~s}^{-1}\) (D) \(3.18 \times 10^{15} \mathrm{~s}^{-1}\)
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