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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

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 ration of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature \(27^{\circ}\) and \(127^{\circ} \mathrm{C}\) respectively is \(\ldots \ldots \ldots\) (A) \((1 / 2)\) (B) \(\sqrt{(3 / 8)}\) (C) \(\sqrt{(8 / 3)}\) (D) 1

A proton and an \(\propto\) -particle are passed through same potential difference. If their initial velocity is zero, the ratio of their de Broglie's wavelength after getting accelerated is \(\ldots \ldots\) (A) \(1: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(2 \sqrt{2}: 1\)

Radius of a nucleus \(2 \times 10^{-15} \mathrm{~m} .\) If we imagine an electron inside the nucleus then energy of electron will be $=\ldots \ldots . \mathrm{MeV}$ $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ (A) \(6.98 \times 10^{3}\) (B) \(8.94 \times 10^{3}\) (C) \(4.98 \times 10^{3}\) (D) \(9.48 \times 10^{3}\)

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) \(941.0 \mathrm{~V}\)
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