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Chapter 17: Problem 2379

de-Broglie wavelength of atom at T \(\mathrm{K}\) absolute temperature will be \(\ldots \ldots \ldots\) (A) \([\mathrm{h} /\\{\mathrm{mkT}\\}]\) (B) \([\mathrm{h} /\\{\sqrt{3} \mathrm{mKT}\\}]\) (C) \([\\{\sqrt{2} \mathrm{mKT}\\} / \mathrm{h}]\) (D) \(\sqrt{(2 \mathrm{mKT})}\)

Short Answer

Expert verified
The short answer for the de-Broglie wavelength of an atom at T K absolute temperature is given by: \[λ = \left[\frac{h}{\sqrt{3mKT}}\right]\] where λ is the de-Broglie wavelength, h is the Planck's constant, m is the mass of the atom, K is the Boltzmann constant, and T is the temperature in Kelvin. So, the correct option is (B).

Step by step solution

01

Calculate the momentum of the atom

Let's start by finding the momentum of the atom with a given kinetic energy. The relation between momentum (p) and kinetic energy (K.E.) is given by the formula: \[K.E. = \frac{p^2}{2m}\] Now, let's plug in the kinetic energy in terms of temperature: \[\frac{3}{2}kT = \frac{p^2}{2m}\]
02

Solve for the momentum (p)

Next, let's solve for the momentum (p) in the above equation: \(p^2 = 3mkT\)
03

Calculate the de-Broglie wavelength (λ)

Now that we have the expression for momentum, let's use the de-Broglie wavelength formula: \[λ = \frac{h}{p}\] Plug in the expression for momentum: \[λ = \frac{h}{\sqrt{3mkT}}\] The above expression matches the option (B) in the given exercise: \[λ = \left[\frac{h}{\sqrt{3mKT}}\right]\] Therefore, the correct answer is (B).

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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) \(43.27 \times 10^{-2} \mathrm{~V}\)

Power produced by a star is \(4 \times 10^{28} \mathrm{~W}\). If the average wavelength of the emitted radiations is considered to be \(4500 \AA\) the number of photons emitted in \(1 \mathrm{~s}\) is \(\ldots \ldots\) (A) \(1 \times 10^{45}\) (B) \(9 \times 10^{46}\) (C) \(8 \times 10^{45}\) (D) \(12 \times 10^{46}\)

An electron is accelerated between two points having potential $20 \mathrm{~V}\( and \)40 \mathrm{~V}$, de-Broglic wavelength of electron is \(\ldots \ldots\) (A) \(0.75 \AA\) (B) \(7.5 \AA\) (C) \(2.75 \AA\) (D) \(0.75 \mathrm{~nm}\)

Which of the following phenomenon can not be explained by quantum theory of light? (A) Emission of radiation from black body (B) Photo electric effect (C) Polarization (D) Crompton effect

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (A) Planck's theory of quantum (p) Light energy \(=\mathrm{hv}\) (B) Einstein's theory of quanta (q) Angular momentum of electron in an orbit. (C) Bohr's stationary orbit (r) Oscillator energies (D) D-Broglie waves (s) Electron microscope (A) \((\mathrm{A}-\mathrm{p}),(\mathrm{b}-\mathrm{q}),(\mathrm{C}-\mathrm{r}),(\mathrm{D}-\mathrm{s})\) (B) \((\mathrm{A}-\mathrm{q}),(\mathrm{B}-\mathrm{r}),(\mathrm{C}-\mathrm{s}),(\mathrm{D}-\mathrm{p})\) (C) \((\mathrm{A}-\mathrm{r}),(\mathrm{B}-\mathrm{p}),(\mathrm{C}-\mathrm{q}),(\mathrm{D}-\mathrm{s})\) (D) \((\mathrm{A}-\mathrm{r}),(\mathrm{B}-\mathrm{p}),(\mathrm{C}-\mathrm{s}),(\mathrm{D}-\mathrm{q})\)
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