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Chapter 17: Problem 2341

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) \(1.826\)

Short Answer

Expert verified
The ratio of de-Broglie wavelengths of a nitrogen gas atom at 300K and 1000K is approximately 1.826, which corresponds to option (D).

Step by step solution

01

Recall de-Broglie wavelength formula

To find the de-Broglie wavelength for an atom of nitrogen, we'll use the de-Broglie wavelength formula, which is: \[ \lambda = \frac{h}{mv} \] Here, \(\lambda\) represents the de-Broglie wavelength, \(h\) denotes Planck's constant (\(6.63 \times 10^{-34}\) Js), \(m\) is the mass of the particle, and \(v\) signifies its velocity. To find the de-Broglie wavelength in both cases, we must determine the necessary velocities of nitrogen atoms at temperatures of 300 K and 1000 K.
02

Determine nitrogen atom velocities using Ideal Gas Law

Beyond the mass, we need the velocities of nitrogen atoms to calculate their de-Broglie wavelengths. We can find those velocities using the Ideal Gas Law: \[ PV = nRT \] Since we have only one atom of nitrogen to work with instead of a full mole, we divide both sides by Avogadro's number (\(N_A\)): \[ \frac{P}{N_A}V = \frac{nR}{N_A}T \] The left side is then simplified to the Boltzmann constant, \(k\), and the right side becomes: \[ \frac{P}{N_A}V = kT \] Velocity relies upon temperature. At a constant pressure of 1 atm, the root mean square (rms) velocity can be defined as: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] With this formula, we can determine the velocity of the nitrogen atom at 300 K and 1000 K.
03

Calculate the de-Broglie wavelength and ratio

Using the previously mentioned de-Broglie wavelength and rms velocity formulas, we can compute the wavelengths at 300 K and 1000 K, respectively. \[ \lambda_{300} = \frac{h}{m v_{rms_{300}}} = \frac{h}{m \sqrt{\frac{3kT_{300}}{m}}} \] \[ \lambda_{1000} = \frac{h}{m v_{rms_{1000}}} = \frac{h}{m \sqrt{\frac{3kT_{1000}}{m}}} \] Now, we'll calculate the ratio of those two wavelengths: \[ \text{Ratio} = \frac{\lambda_{300}}{\lambda_{1000}} = \frac{\frac{h}{m \sqrt{\frac{3kT_{300}}{m}}}}{\frac{h}{m \sqrt{\frac{3kT_{1000}}{m}}}} \] Simplify and solve for the ratio: \[ \text{Ratio} = \frac{\sqrt{\frac{3kT_{1000}}{m}}}{\sqrt{\frac{3kT_{300}}{m}}} = \sqrt{\frac{T_{1000}}{T_{300}}} \] \[ = \sqrt{\frac{1000}{300}} = \sqrt{\frac{10}{3}} \approx 1.826 \] Therefore, the ratio of de-Broglie wavelengths of a nitrogen gas atom at 300K and 1000K is approximately 1.826, which corresponds to option (D).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a fundamental equation that describes the behavior of an ideal gas. It is expressed as: \[PV = nRT\]- **P**: Pressure of the gas- **V**: Volume of the gas- **n**: Amount of substance (in moles)- **R**: Universal gas constant (8.314 J/mol K)- **T**: Temperature in KelvinThis equation helps us understand how a gas behaves under different conditions. By rearranging the Ideal Gas Law, we can calculate various properties of the gas when given certain variables. For instance, in the exercise provided, it is the basis for deriving the velocity of the nitrogen atom. Under certain conditions, like when dealing with a single atom, knowing the volume is not required, and the focus shifts to deriving other important quantities like kinetic energy and velocity using related concepts.
Root Mean Square Velocity
Root mean square (rms) velocity is a measure of the speed of particles in a gas, crucial for determining transport properties. It is given by the equation: \[v_{rms} = \sqrt{\frac{3kT}{m}}\]- **k**: Boltzmann constant - **T**: Temperature in Kelvin- **m**: Mass of a single particleIn the context of gases, rms velocity provides an average kinetic energy of the particles since not all particles move at the same speed. The rms velocity increases with temperature, directly influencing parameters like de-Broglie wavelength. For the nitrogen atom at different temperatures, calculating the rms velocity allows us to find its de-Broglie wavelength 鈥� a crucial step as shown in the provided exercise solution.
Temperature Dependence
Temperature plays a pivotal role in gas behavior and directly affects properties like velocity and pressure. According to kinetic molecular theory, increasing temperature results in increased energy within gas particles, leading to faster movement. In formulas, temperature is always included in its absolute scale, Kelvin, to ensure accuracy as Celsius or Fahrenheit does not behave proportionally in calculations. The exercise exemplifies how temperature affects the de-Broglie wavelength of nitrogen by demonstrating that varying temperature leads to changing particle velocity. The higher the temperature, the higher the rms velocity, thus influencing the wavelength ratio 鈥� a key insight in understanding gas physics.
Planck's Constant
Planck's constant is a physical constant central to quantum mechanics and is crucial in calculating de-Broglie wavelengths. It is denoted as \(h = 6.63 \times 10^{-34}\) Js and represents the quantization of energy levels.In the de-Broglie formula \(\lambda = \frac{h}{mv}\), Planck's constant allows us to relate a particle's velocity to its wavelength, which is especially significant at atomic and subatomic scales.Understanding Planck's constant offers a bridge between classical physics and quantum mechanics, showing how even macroscopic objects have wave characteristics as seen through de-Broglie's hypothesis.

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Most popular questions from this chapter

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (I) Quantization of charge (P) Diffraction of light (II) Wave nature of light (Q) de Broglie hypothesis (III) Dual nature of matter (R) Photo-electric effect (IV) Particle nature of light (S) Millikan's drop experiment (A) \(\mathrm{I}-\mathrm{P}, \mathrm{II}-\mathrm{Q}, \mathrm{III}-\mathrm{R}, \mathrm{IV}-\mathrm{S}\) (B) \(\mathrm{I}-\mathrm{S}, \mathrm{II}-\mathrm{P}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{R}\) (C) \(\mathrm{I}-\mathrm{Q}, \mathrm{II}-\mathrm{R}, \mathrm{III}-\mathrm{S}, \mathrm{IV}-\mathrm{P}\) (D) \(I-R . I I-S . I I I-P . I V-O\)

For wave concerned with proton, de-Broglie wavelength change by \(0.25 \%\). If its momentum changes by \(\mathrm{P}_{\mathrm{O}}\) initial momentum \(=\ldots \ldots \ldots\) (A) \(100 \mathrm{P}_{\mathrm{O}}\) (B) \(\left\\{\mathrm{P}_{\mathrm{O}} / 400\right\\}\) (C) \(401 \mathrm{P}_{\mathrm{O}}\) (D) \(\left\\{\mathrm{P}_{\mathrm{O}} / 100\right\\}\)

Direction Read the following question choose if: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) Assertion is true but the Reason is false. (d) Both Assertion and Reason is false. Assertion : The de Broglie wavelength of an electron accelerated through 941 volts is \(0.4 \AA\). Reason: Higher the acceleration potentials of electron, smaller is the de Broglie wavelength. (A) a (B) \(\mathrm{b}\) (C) (D) \(\mathrm{d}\)

Output power of He-Ne LASER of low energy is \(1.00 \mathrm{~mW}\). Wavelength of the light is \(632.8 \mathrm{~nm}\). What will be the number of photons emitted per second from this LASER? (A) \(8.31 \times 10^{15} \mathrm{~s}^{-1}\) (B) \(5.38 \times 10^{15} \mathrm{~s}^{-1}\) (C) \(1.83 \times 10^{15} \mathrm{~s}^{-1}\) (D) \(3.18 \times 10^{15} \mathrm{~s}^{-1}\)

Frequency of incident light on body is \(\mathrm{f}\). Threshold frequency of body is \(f_{0}\). Maximum velocity of electron \(=\ldots \ldots \ldots\).. where \(m\) is mass of electron. (A) \(\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]^{(1 / 2)}\) (B) \(\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]\) (C) \([2 \mathrm{hf} / \mathrm{m}]^{(1 / 2)}\) (D) \(\mathrm{h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\)
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