What is the ratio of De-Broglie wavelengths for helium to hydrogen at room temperature?

What is the ratio of De-Broglie wavelengths for helium to hydrogen at room temperature?

• A. 1:1
• B. 3:1
• C. 2:1
• D. 0.5:1

Answer: (A)

The ratio of De-Broglie wavelengths for two gases can be derived from the formula for the De-Broglie wavelength, which is given by: \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the De-Broglie wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle. The momentum \( p \) of a particle is related to its mass \( m \) and velocity \( v \) by the equation: \[ p = mv \] At a given temperature, the average kinetic energy of gas particles is related to the temperature by: \[ KE = \frac{3}{2} kT \] where \( k \) is Boltzmann's constant and \( T \) is the temperature in Kelvin. The mean speed \( v \) of gas molecules is also dependent on their molar mass \( M \): \[ v = \sqrt{\frac{3kT}{M}} \] Combining these equations, we find that the De-Broglie wavelength can be expressed in terms of mass: \[ \lambda \propto \frac{1}{

The ratio of De-Broglie wavelengths for two gases can be derived from the formula for the De-Broglie wavelength, which is given by:

[

\lambda = \frac{h}{p}

]

where ( \lambda ) is the De-Broglie wavelength, ( h ) is Planck's constant, and ( p ) is the momentum of the particle. The momentum ( p ) of a particle is related to its mass ( m ) and velocity ( v ) by the equation:

[

p = mv

]

At a given temperature, the average kinetic energy of gas particles is related to the temperature by:

[

KE = \frac{3}{2} kT

]

where ( k ) is Boltzmann's constant and ( T ) is the temperature in Kelvin. The mean speed ( v ) of gas molecules is also dependent on their molar mass ( M ):

[

v = \sqrt{\frac{3kT}{M}}

]

Combining these equations, we find that the De-Broglie wavelength can be expressed in terms of mass:

[

\lambda \propto \frac{1}{