Question

The atomic masses of He and Ne are 4 and 20 amu respectively. The value of the de Broglie wavelength of He gas at -73°C is ‘M’ times that of the de Broglie wavelength of Ne at 727°C ‘M’ is.

Answer 1

de Broglie’s wavelength of a particle when kinetic energy (K.E) and mass (m) are given:

 

λ = h / √2K.E m

Given:

Mass of He atom (mHe) = 4 amu

Mass of He atom (mNe) = 20 amu

The temperature of He(THe) = -73°C = 200K

The temperature of Ne(TNe) = +727°C = 1000K

 

We know that;

K.E ∝ T

K.EHe / K.ENe = THe / TNe = 200 / 1000 = ⅕

Now, the ratio of de Broglie’s wavelengths of Ne and He (λHe / λNe)

λHe / λNe = √2K.ENemNe / √2K.EHemHe = √ 5 / 1 x 20 / 4 = 5

∴ λHe = 5 x λNe

 

The value of m is 5.

 

 

The value of m is 5.