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Chapter 17: Problem 2418

The work function of metal is \(5.3 \mathrm{eV}\). What is threshold frequency? (A) \(3.1 \times 10^{15} \mathrm{~Hz}\) (B) \(3.1 \times 10^{45} \mathrm{~Hz}\) (C) \(1.3 \times 10^{15} \mathrm{~Hz}\) (D) \(1.3 \times 10^{45} \mathrm{~Hz}\)

Short Answer

Expert verified
The threshold frequency can be calculated using the photoelectric effect equation \(ϕ = h \times f\), where \(ϕ\) is the work function and \(h\) is Planck's constant. Converting the given work function \(5.3 \mathrm{eV}\) to Joules, we get \(8.48 \times 10^{-19} \mathrm{J}\). Using Planck's constant \(h = 6.63 \times 10^{-34} \mathrm{Js}\), we can find the threshold frequency \(f = \frac{ϕ}{h} = 1.28 \times 10^{15} \mathrm{Hz}\). Comparing with the options, the closest option is (C) \(1.3 \times 10^{15} \mathrm{Hz}\).

Step by step solution

01

Write down the photoelectric effect equation

The photoelectric effect equation relating the work function (ϕ), Planck's constant (h), and the threshold frequency (f) is: \(ϕ = h \times f\)
02

Identify the given work function and write it in Joules

The given work function is 5.3 eV. We need to convert it to Joules using the following relationship: 1 eV = \(1.6 \times 10^{-19} J\) So, the work function in Joules is: \(ϕ = 5.3 \mathrm{eV} \times 1.6 \times 10^{-19} \frac{\mathrm{J}}{\mathrm{eV}} = 8.48 \times 10^{-19} \mathrm{J}\)
03

Find Planck's constant

Planck's constant (h) has a fixed value, which is: \(h = 6.63 \times 10^{-34} \mathrm{Js}\)
04

Calculate the threshold frequency

Now we can use the photoelectric effect equation to find the threshold frequency: \(f = \frac{ϕ}{h}\) \(f = \frac{8.48 \times 10^{-19} \mathrm{J}}{6.63 \times 10^{-34} \mathrm{Js}}\) \(f = 1.28 \times 10^{15} \mathrm{Hz}\)
05

Match the calculated threshold frequency with the given options

Now that we have calculated the threshold frequency, we can compare it to the given options: (A) \(3.1 \times 10^{15} \mathrm{~Hz}\) (B) \(3.1 \times 10^{45} \mathrm{~Hz}\) (C) \(1.3 \times 10^{15} \mathrm{~Hz}\) (D) \(1.3 \times 10^{45} \mathrm{~Hz}\) The calculated threshold frequency \(1.28 \times 10^{15} \mathrm{Hz}\) is approximately equal to (C) \(1.3 \times 10^{15} \mathrm{Hz}\). Therefore, the correct answer is (C).

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Most popular questions from this chapter

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

The de-Broglie wavelength associated with a particle with rest mass \(\mathrm{m}_{0}\) and moving with speed of light in vacuum is..... (A) \(\left(\mathrm{h} / \mathrm{m}_{0} \mathrm{c}\right)\) (B) 0 (C) \(\infty\) (D) \(\left(\mathrm{m}_{0} \mathrm{c} / \mathrm{h}\right)\)

A proton and electron are lying in a box having impenetrable walls, the ratio of uncertainty in their velocities are \(\ldots \ldots\) \(\left(\mathrm{m}_{\mathrm{e}}=\right.\) mass of electron and \(\mathrm{m}_{\mathrm{p}}=\) mass of proton. (A) \(\left(\mathrm{m}_{\mathrm{e}} / \mathrm{m}_{\mathrm{p}}\right)\) (B) \(\mathrm{m}_{\mathrm{e}} \cdot \mathrm{m}_{\mathrm{p}}\) (C) \(\left.\sqrt{\left(m_{e}\right.} \cdot m_{p}\right)\) (D) \(\sqrt{\left(m_{e} / m_{p}\right)}\)

Select the correct statement from the following (A) Radiation and matter (particles) may not exhibit both the wave nature and particle nature simultaneously at the some moment. (B) At some moment electromagnetic waves get divided in to small pieces named particles (C) In a given circumstance a particle at one moment behaves like particle and the at the next moment as wave and so on (D) Each microscopic particle is enveloped by a wave

A photon having energy \(5.5 \mathrm{eV}\) is incident on metal suface and emits photo electrons having maximum kinetic energy \(0.4 \mathrm{eV}\). Then stopping potential of this electron is..... (A) \(5.5 \mathrm{eV}\) (B) \(5.1 \mathrm{eV}\) (C) \(5.9 \mathrm{eV}\) (D) \(0.4 \mathrm{eV}\)
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