Question

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

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

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

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: \(\lambda = \dfrac{6.625 \times 10^{-34} Js}{(1.6 \times 10^{-27} kg)(100 m/s)}\)

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) \(3.96 \AA\)