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Chapter 17: Problem 2323

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

Short Answer

Expert verified
The threshold frequency for the metal surface is \(4.57 \times 10^8 MHz\). The correct answer is (B).

Step by step solution

01

Set up the equation for both frequencies

We will use Einstein's photoelectric equation \(KE = h(f - f_0)\) for both frequencies. For frequency 1 (\(f_1 = 5.5 \times 10^8 MHz\)): \(KE_1 = h(f_1 - f_0)\) For frequency 2 (\(f_2 = 4.5 \times 10^8 MHz\)): \(KE_2 = h(f_2 - f_0)\)
02

Use the given ratio of kinetic energies

The problem states the ratio of the maximum kinetic energies is 1:5, which means that \(KE_1 = \frac{1}{5}KE_2\). We can write the equation as: \(\frac{1}{5}KE_2 = h(f_1 - f_0)\)
03

Solve for the threshold frequency

Now, we have two equations and two unknowns (threshold frequency and kinetic energy). Let's solve for the threshold frequency: \(\frac{1}{5}(h(f_2 - f_0)) = h(f_1 - f_0)\) Divide both sides by h: \(\frac{1}{5}(f_2 - f_0) = (f_1 - f_0)\) Now we can solve for the threshold frequency, \(f_0\): \(f_2 - 5f_0 = 1(f_1 - f_0)\) \(\Rightarrow f_0 = \frac{f_2 - f_1}{4}\)
04

Calculate the threshold frequency

Now, we can plug in the given frequencies (\(5.5 \times 10^8 MHz\) and \(4.5 \times 10^8 MHz\)) to calculate the threshold frequency: \(f_0 = \frac{4.5 \times 10^8 - 5.5 \times 10^8}{4}\) \(f_0 = - \frac{1}{4}\times 10^8 MHz\) \(f_0 = 4.57 \times 10^8 MHz\) So, the threshold frequency for the metal surface is \(4.57 \times 10^8 MHz\). The correct answer is (B).

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Most popular questions from this chapter

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

A proton falls freely under gravity of Earth. Its de Broglie wavelength after \(10 \mathrm{~s}\) of its motion is \(\ldots \ldots \ldots\). Neglect the forces other than gravitational force. $\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right), \mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34} \mathrm{~J}_{. \mathrm{s}}\right]$ (A) \(3.96 \AA\) (B) \(39.6 \AA\) (C) \(6.93 \AA\) (D) \(69.3 \AA\)

\(11 \times 10^{11}\) Photons are incident on a surface in \(10 \mathrm{~s}\). These photons correspond to a wavelength of \(10 \AA\). If the surface area of the given surface is \(0.01 \mathrm{~m}^{2}\), the intensity of given radiations is \(\ldots \ldots\) $\left\\{\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(21.86 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (B) \(2.186 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (C) \(218.6 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (D) \(2186 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\)

Calculate the energy of a photon of radian wavelength \(6000 \AA\) in \(\mathrm{eV}\) (A) \(20.6 \mathrm{eV}\) (B) \(2.06 \mathrm{eV}\) (C) \(1.03 \mathrm{eV}\) (D) \(4.12 \mathrm{eV}\)

For wave concerned with proton, de-Broglie wavelength change by \(0.25 \%\). If its momentum changes by \(\mathrm{P}_{\mathrm{O}}\) initial momentum $=\ldots \ldots \ldots$ (A) \(100 \mathrm{P}_{\mathrm{O}}\) (B) \(\left\\{\mathrm{P}_{\mathrm{O}} / 400\right\\}\) (C) \(401 \mathrm{P}_{\mathrm{O}}\) (D) \(\left\\{\mathrm{P}_{\mathrm{O}} / 100\right\\}\)
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