Question

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

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

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.

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.

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).