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Chapter 17: Problem 2329

At \(10^{\circ} \mathrm{C}\) temperature, de-Broglie wave length of atom is $0.4 \AA\(.If temperature of atom is increased by \)30^{\circ} \mathrm{C}$, what will be change in de-Broglie wavelength of atom? (A) decreases \(10^{-2} \AA\) (B) decreases \(2 \times 10^{-2} \AA\) (C) increases \(10^{-2} \AA\) (D) increases \(2 \times 10^{-2} \AA\)

Short Answer

Expert verified
The change in de-Broglie wavelength, \(\Delta\lambda\), is equal to an increase of \(2 \times 10^{-2} \AA.\) The correct answer is (D).

Step by step solution

01

Write the de-Broglie wavelength formula for an atom

The de-Broglie wavelength formula for an atom can be given as follows: \( \lambda = \frac{h}{p} \) where \(\lambda\) is the de-Broglie wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the atom.
02

Write the temperature-energy relation for an atom

The temperature has an impact on the kinetic energy of an atom, which in turn impacts its momentum. The average kinetic energy of an atom can be expressed in terms of temperature as follows: \( \text{KE} = \frac{3}{2}k_\mathrm{B}T \) where \(\text{KE}\) is the average kinetic energy of an atom, \(k_\mathrm{B}\) is the Boltzmann constant, and \(T\) is the temperature in Kelvin.
03

Calculate the initial momentum and wavelength of the atom

We are given that the initial de-Broglie wavelength of the atom is \(0.4\, \text{\AA}\) or \(4 \times 10^{-11}\, \text{m}\) at \(10^{\circ} \mathrm{C}\) or \(283.15\, \text{K}\). We can use the de-Broglie wavelength formula and temperature-energy relation to find the initial momentum of the atom. From the de-Broglie wavelength formula, we can get the momentum as: \( p = \frac{h}{\lambda} \) Since the average kinetic energy of an atom is related to temperature as mentioned in Step 2, we can use that relation to find the initial momentum: \( p^2 = 2m(\text{KE}) \)
04

Calculate the final momentum and wavelength of the atom

We are given that the temperature is increased by \(30^{\circ} \mathrm{C}\), i.e., the final temperature is \((10 + 30)^{\circ} \mathrm{C}\) or \(313.15\, \text{K}\). We can use the temperature-energy relation to find the final momentum of the atom and then use the de-Broglie wavelength formula to find the final wavelength of the atom.
05

Calculate the change in de-Broglie wavelength

After finding the initial and final wavelengths of the atom, we can calculate the change in de-Broglie wavelength by subtracting the initial wavelength from the final wavelength: \( \Delta\lambda = \lambda_\mathrm{final} - \lambda_\mathrm{initial} \) By calculating the value of \(\Delta\lambda\), we can determine which option among (A), (B), (C), and (D) is correct, considering that each option states whether the change in de-Broglie wavelength results in an increase or decrease and by what magnitude.

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Most popular questions from this chapter

The de-Broglie wavelength associated with a particle with rest mass \(\mathrm{m}_{0}\) and moving with speed of light in vacuum is..... (A) \(\left(\mathrm{h} / \mathrm{m}_{0} \mathrm{c}\right)\) (B) 0 (C) \(\infty\) (D) \(\left(\mathrm{m}_{0} \mathrm{c} / \mathrm{h}\right)\)

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) \(1.826\)

Energy corresponding to threshed frequency of metal is \(6.2 \mathrm{eV}\). If stopping potential corresponding to radiation incident on surface is $5 \mathrm{~V}\(, incident radiation will be in the \)\ldots \ldots \ldots \ldots \ldots$ region. (A) X-ray (B) Ultraviolet (C) infrared (D) Visible

Uncertainty in position of electron is found of the order of de-Broglie wavelength. Using Heisenberg's uncertainty principle, it is found that order of uncertainty in its velocity \(=\ldots \ldots \ldots .\) (A) \(1 \mathrm{v}\) (B) \(2 \mathrm{v}\) (C) \((\mathrm{v} / 2 \pi)\) (D) \(2 \pi \mathrm{v}\)

The cathode of a photoelectric cell is changed such that the work function changes from \(\mathrm{W}_{1}\) to \(\mathrm{W}_{2}\left(\mathrm{~W}_{2}>\mathrm{W}_{1}\right)\). If the currents before and after change are \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\), all other conditions remaining unchanged, then assuming $\mathrm{hf}>\mathrm{W}_{2} \ldots \ldots$ (A) \(\mathrm{I}_{1}=\mathrm{I}_{2}\) (B) \(I_{1}\mathrm{I}_{2}\) (D) \(\mathrm{I}_{1}<\mathrm{I}_{2}<2 \mathrm{I}_{1}\)
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