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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

An electron is accelerated under a potential difference of \(64 \mathrm{~V}\), the de Broglie wave length associated with electron is $=\ldots \ldots \ldots \ldots . \AA$ (Use charge of election $=1.6 \times 10^{-19} \mathrm{C},=9.1 \times 10^{-31} \mathrm{Kg}$, mass of electrum \(\mathrm{h}=6.6623 \times 10^{-43} \mathrm{~J}\).sec \()\) (A) \(4.54\) (B) \(3.53\) (C) \(2.53\) (D) \(1.534\)

In photoelectric effect, work function of martial is \(3.5 \mathrm{eV}\). By applying \(-1.2 \mathrm{~V}\) potential, photo electric current becomes zero, so......... (A) energy of incident photon is \(4.7 \mathrm{eV}\). (B) energy of incident photon is \(2.3 \mathrm{eV}\) (C) If photon having higher frequency is used, photo electric current is produced. (D) When energy of photon is \(2.3 \mathrm{eV}\), photo electric current becomes maximum

Ration of momentum of photons having wavelength \(4000 \AA \& 8000 \AA\) is ........... (A) \(2: 1\) (B) \(1: 2\) (C) \(20: 1\) (D) \(1: 20\)

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) \(69.3 \AA\)

The work function for tungsten and solidus are \(4.5 \mathrm{eV}\) and $2.3 \mathrm{eV}\( respectively. If the threshold wavelength \)\lambda_{0}$ for sodium is \(5460 \AA\) the value of \(\lambda_{0}\) for tungsten is $\ldots \ldots$ (A) \(528 \AA\) (B) \(2791 \AA\) (C) \(589 \AA\) (D) \(10683 \AA\)
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