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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Threshold Frequency
The threshold frequency is a critical concept in understanding the photoelectric effect. It is the minimum frequency that a photon should have to eject electrons from a metal surface. The photoelectric effect equation relates frequency to the energy of incident photons: \(E = h u = \phi + KE\). When kinetic energy \(KE\) is zero, the equation simplifies to \(h u = \phi\). For every metal, there is a unique threshold frequency based on its atomic structure.
  • If a photon鈥檚 frequency is below this threshold, no electrons will be emitted, regardless of its intensity.
  • If the frequency meets or exceeds this threshold, electrons will be emitted.
Understanding this concept helps in examining how different materials interact with light.
Work Function
The work function \(\phi\) refers to the minimum energy needed to remove an electron from a solid surface. Different materials have different work functions based on their electronic properties. It is deeply connected with the threshold frequency as expressed in the relation \(\phi = h u\).As we see with the formula, the work function is measured in energy units, typically electronvolts (eV) or Joules (J).
  • This energy corresponds to the threshold frequency through the conversion factor \(h\), Planck鈥檚 constant.
  • In the exercise, the ratio of threshold frequencies points directly to the ratio of work functions, since \(u_1 / u_2 = 1/3\) led to \(\phi_1 / \phi_2 = 1/3\).
Understanding the work function helps us predict and explain how metals respond to different frequencies of light.
Planck's Constant
Planck's constant \(h\) is a fundamental constant in quantum mechanics, instrumental in linking frequency and energy in the photoelectric effect.Its value is approximately \(6.626 \times 10^{-34} \text{J s}\) and is crucial in equations dealing with quantum phenomena.
  • In the photoelectric equation \(E = h u\), \(h\) serves as a proportionality constant between energy \(E\) and frequency \(u\).
  • It predicts observable phenomena in particle emission, like those seen with different metals, and is central to understanding quantum mechanics.
Knowledge of Planck's constant allows us to quantify the interaction between matter and light on extremely small scales.

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Most popular questions from this chapter

The cathode of a photoelectric cell is changed such that the work function changes from \(\mathrm{W}_{1}\) to \(\mathrm{W}_{2}\left(\mathrm{~W}_{2}>\mathrm{W}_{1}\right)\). If the currents before and after change are \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\), all other conditions remaining unchanged, then assuming \(\mathrm{hf}>\mathrm{W}_{2} \ldots \ldots\) (A) \(\mathrm{I}_{1}=\mathrm{I}_{2}\) (B) \(I_{1}\mathrm{I}_{2}\) (D) \(\mathrm{I}_{1}<\mathrm{I}_{2}<2 \mathrm{I}_{1}\)

If intensity of incident light is increased, ........of photo electrons will increase. (A) number (B) Frequency (C) energy (D) wavelength

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) \(1.7 \times 10^{7} \mathrm{sec}\)

Energy corresponding to threshed frequency of metal is \(6.2 \mathrm{eV}\). If stopping potential corresponding to radiation incident on surface is \(5 \mathrm{~V}\), incident radiation will be in the \(\ldots \ldots \ldots \ldots \ldots\) region. (A) X-ray (B) Ultraviolet (C) infrared (D) Visible

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) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)
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