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Chapter 17: Problem 2373

Frequency of photon having energy \(66 \mathrm{eV}\) is ....... \(\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)\) (A) \(8 \times 10^{-15} \mathrm{~Hz}\) (B) \(12 \times 10^{-15} \mathrm{~Hz}\) (C) \(16 \times 10^{-15} \mathrm{~Hz}\) (D) \(24 \times 10^{+15} \mathrm{~Hz}\)

Short Answer

Expert verified
The frequency of the photon having energy 66 eV is \(16 \times 10^{-15}\mathrm{~Hz}\).

Step by step solution

01

Convert energy to joules

Given energy, E = 66 eV. To convert energy into joules, we can use the conversion factor 1 eV = 1.6 × 10^{-19} J. \(E = 66 \text{ eV} \times \frac{1.6 \times 10^{-19} J}{1 \text{ eV}}\) E = 1.056 × 10^{-17} J Now we have the energy in joules.
02

Use the formula to find frequency

Now we can use the formula: Energy (E) = Planck's constant (h) × Frequency (ν) We are given Planck's constant, h = 6.6 × 10^{-34} J·s. We can rearrange the formula to find the frequency: \(ν = \frac{E}{h}\) Plug in the values for E and h: \(ν = \frac{1.056 \times 10^{-17} J}{6.6 \times 10^{-34} J.s}\)
03

Calculate the frequency and select the correct option

After dividing the energy by Planck's constant, we get the frequency: ν = 1.6 × 10^{16} Hz We can write this in scientific notation as: ν = 16 × 10^{15} Hz Comparing this value to the available options, we see that it matches option (C): (C) \(16 \times 10^{-15} \mathrm{~Hz}\) So the frequency of the photon having energy 66 eV is \(16 \times 10^{-15}\mathrm{~Hz}\). Option (C) is the correct answer.

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Most popular questions from this chapter

The work function of a metal is \(1 \mathrm{eV}\). Light of wavelength $3000 \AA$ is incident on this metal surface. The maximum velocity of emitted photoelectron will be \(\ldots \ldots .\) (A) \(10 \mathrm{~ms}^{-1}\) (B) \(10^{3} \mathrm{~ms}^{-1}\) (C) \(10^{4} \mathrm{~ms}^{-1}\) (D) \(10^{6} \mathrm{~ms}^{-1}\)

Wavelength \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\) are incident on two identical metal plates and photo electrons are emitted. If \(\lambda_{\mathrm{A}}=2 \lambda_{\mathrm{B}}\), the maximum kinetic energy of photo electrons is \(\ldots \ldots \ldots\) (A) \(2 \mathrm{~K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}\) (B) \(\mathrm{K}_{\mathrm{A}}<\left(\mathrm{K}_{\mathrm{B}} / 2\right)\) (C) \(\mathrm{K}_{\mathrm{A}}=2 \mathrm{~K}_{\mathrm{B}}\) (D) \(\mathrm{K}_{\mathrm{A}}>\left(\mathrm{K}_{\mathrm{B}} / 2\right)\)

Find the velocity at which mass of a proton becomes \(1.1\) times its rest mass, \(\mathrm{m}_{\mathrm{g}}=1.6 \times 10^{-27} \mathrm{~kg}\) Also, calculate corresponding temperature. For simplicity, consider a proton as non- interacting ideal-gas particle at \(1 \mathrm{~atm}\) pressure. $\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \cdot \mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) $\mathrm{V}=1.28 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{12} \mathrm{~K}$ (B) $\mathrm{V}=12.6 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{11} \mathrm{~K}$ (C) $\mathrm{V}=1.26 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=5.76 \times 10^{11} \mathrm{~K}$ (D) $\mathrm{V}=12.6 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.56 \times 10^{11} \mathrm{~K}$

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) \(3.28 \mathrm{~V}\)

Energy of photon of light having two different frequencies are \(2 \mathrm{eV}\) and \(10 \mathrm{eV}\) respectively. If both are incident on the metal having work function \(1 \mathrm{eV}\), ratio of maximum velocities of emitted electron is ............ (A) \(1: 5\) (B) \(3: 11\) (C) \(2: 9\) (D) \(1: 3\)
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