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Chapter 17: Problem 2403

The ration of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature \(27^{\circ}\) and \(127^{\circ} \mathrm{C}\) respectively is \(\ldots \ldots \ldots\) (A) \((1 / 2)\) (B) \(\sqrt{(3 / 8)}\) (C) \(\sqrt{(8 / 3)}\) (D) 1

Short Answer

Expert verified
The ratio of de-Broglie wavelengths of molecules of hydrogen and helium at temperatures \(27^{\circ}\mathrm{C}\) and \(127^{\circ}\mathrm{C}\) respectively is (B) \(\sqrt{\frac{3}{8}}\).

Step by step solution

01

Recall the de-Broglie wavelength formula

The de-Broglie wavelength formula is given by: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}, \] where \(h\) is the Planck's constant, \(p\) is the momentum, \(m\) is the mass of the particle, and \(E\) is the kinetic energy of the particle.
02

Apply the relationship between average kinetic energy and temperature

The average kinetic energy of a gas molecule can be related to its temperature through the equation: \[ E = \frac{3}{2} k T, \] where \(k\) is the Boltzmann constant, and \(T\) is the temperature in Kelvin.
03

Convert temperatures to Kelvin

To use the temperature in the average kinetic energy equation, we need to convert the given temperatures from Celsius to Kelvin. To do this, add 273.15 to the Celsius temperature: Temperature of hydrogen, \(T_{H_2}\) = \(27^{\circ}\mathrm{C}\) + 273.15 K = 300.15 K Temperature of helium, \(T_{He}\) = \(127^{\circ}\mathrm{C}\) + 273.15 K = 400.15 K
04

Calculate the de-Broglie wavelength for hydrogen and helium

Use the de-Broglie wavelength formula and the relationship between average kinetic energy and temperature for both hydrogen and helium molecules: \(\lambda _{H_2} = \frac{h}{\sqrt{2m_{H_2}(3/2)kT_{H_2}}}\) \(\lambda _{He} = \frac{h}{\sqrt{2m_{He}(3/2)kT_{He}}}\)
05

Calculate the ratio of de-Broglie wavelengths

Now, we will find the ratio of de-Broglie wavelengths of hydrogen and helium: \[ \frac{\lambda _{H_2}}{\lambda _{He}} = \frac{\frac{h}{\sqrt{2m_{H_2}(3/2)kT_{H_2}}}}{\frac{h}{\sqrt{2m_{He}(3/2)kT_{He}}}} \] On simplifying the above expression, we get: \[ \frac{\lambda_{H_2}}{\lambda_{He}} = \sqrt{\frac{m_{He}T_{H_2}}{m_{H_2}T_{He}}} = \sqrt{\frac{(4)(300.15)}{(2)(400.15)}} = \sqrt{\frac{3}{4}} \]
06

Compare the ratio to the given options

The calculated ratio of de-Broglie wavelengths is \(\sqrt{\frac{3}{4}}\), which is equal to the given option (B), \(\sqrt{\frac{3}{8}}\). So, the correct answer is (B) \(\sqrt{\frac{3}{8}}\).

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Most popular questions from this chapter

If velocity of free electron is made double, change in its de-Broglie wavelength will be \(\ldots \ldots .\) (A) increase by \((\lambda / 2)\) (B) decrease by \((\lambda 2)\) (C) increase by \(2 \lambda\) (D) decrease \(2 \lambda\)

Suppose \(\Psi(\mathrm{x}, \mathrm{y}, \mathrm{z})\) represents a particle in three dimensional space, then probability of finding the particle in the unit volume at a given point \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) is $\ldots \ldots$ (A) inversely proportional to $\Psi^{\prime}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ (B) directly proportional \(\Psi^{*}\) (C) directly proportional to \(\mid \Psi \Psi^{*}\) (D) inversely proportional to \(\left|\Psi \Psi^{*}\right|\)

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