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Chapter 17: Problem 2346

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) \(69.3 \AA\)

Short Answer

Expert verified
The short answer for the de Broglie wavelength of a proton after falling freely for 10 seconds under Earth's gravitational force is approximately (A) \(3.96 \AA\) .

Step by step solution

01

Determine the velocity of the proton after falling for 10 seconds

Using the kinematic equation, we can find the velocity (v) of the proton after 10 seconds as: v = u + gt Where: u = initial velocity = 0 (since the proton is falling freely) g = gravitational acceleration = 10 m/s² t = time = 10 s So, we have: v = 0 + (10)(10) = 100 m/s
02

Calculate the de Broglie wavelength of the proton

The de Broglie wavelength formula is given by: \(\lambda = \dfrac{h}{m_p v}\) Where: \(\lambda\) = de Broglie wavelength h = Planck's constant = \(6.625 \times 10^{-34} J\cdot s\) \(m_p\) = mass of the proton = \(1.6 \times 10^{-27} kg\) v = velocity of the proton = 100 m/s Now, plug in the values to find the de Broglie wavelength: \(\lambda = \dfrac{6.625 \times 10^{-34} Js}{(1.6 \times 10^{-27} kg)(100 m/s)}\)
03

Solve for the de Broglie wavelength

Simplifying the equation, we get: \(\lambda = \dfrac{6.625 \times 10^{-34}}{1.6 \times 10^{-24}}\) \(\lambda = \dfrac{6.625}{1.6} \times 10^{-10} m\) Now let's convert this to Angstrom unit (\(\AA\)): 1 Angstrom (\(\AA\)) = \(10^{-10}m\) \(\lambda = 4.14 \times 10^{-10} m = 4.14 \AA\) Since 4.14 is the closest to 3.96 \(\AA\) (A), we can conclude that the correct answer is: (A) \(3.96 \AA\)

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Most popular questions from this chapter

Wavelength of light incident on a photo-sensitive surface is reduced form \(3500 \AA\) to \(290 \mathrm{~mm}\). The change in stopping potential is $\ldots \ldots . .\left(\mathrm{h}=6.625 \times 10^{-24} \mathrm{~J} . \mathrm{s}\right)$ (A) \(42.73 \times 10^{-2} \mathrm{~V}\) (B) \(27.34 \times 10^{-2} \mathrm{~V}\) (C) \(73.42 \times 10^{-2} \mathrm{~V}\) (D) \(43.27 \times 10^{-2} \mathrm{~V}\)

A proton, a deuteron and an \(\propto\) -particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is (A) \(1: 2: 8\) (B) \(1: 2: 4\) (C) \(1: 1: 2\) (D) None of these

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

De-Broglie wavelength of particle moving at a (1/4) th of speed of light having rest mass \(\mathrm{m}_{0}\) is \(\ldots \ldots \ldots\) (A) $\left\\{(3.87 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (B) $\left\\{(4.92 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (C) $\left\\{(7.57 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (D) $\left\\{(9.46 \mathrm{~h}) /\left(\mathrm{m}_{\circ} \mathrm{C}\right)\right\\}$

What should be the ratio of de-Broglie wavelength of an atom of nitrogen gas at \(300 \mathrm{~K}\) and \(1000 \mathrm{~K}\). Mass of nitrogen atom is $4.7 \times 10^{-26} \mathrm{~kg}$ and it is at 1 atm pressure Consider it as an idecal gas. (A) \(2.861\) (B) \(8.216\) (C) \(6.281\) (D) \(1.826\)
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