Assuming the nitrogen molecule is moving with | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The $r.m.s.$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.The de$-$Broglie wavelength is given by $\lambda = \frac{h}{m

Vedclass · Accepted Answer

$0.24$

Vedclass · Answer

(B) The $r.m.s.$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.The de$-$Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Substituting $v_{rms}$ for $v$, we get $\lambda = \frac{h}{m \sqrt{\frac{3kT}{m}}} = \frac{h}{\sqrt{3kTm}}$.Given values: $h = 6.63 	imes 10^{-34} \ J \cdot s$, $k = 1.38 	imes 10^{-23} \ J/K$, $T = 400 \ K$, and $m = 4.64 	imes 10^{-26} \ kg$.Calculating the denominator: $\sqrt{3 	imes 1.38 	imes 10^{-23} 	imes 400 	imes 4.64 	imes 10^{-26}} = \sqrt{7.68768 	imes 10^{-45}} \approx 2.77 	imes 10^{-22} \ kg \cdot m/s$.Now, $\lambda = \frac{6.63 	imes 10^{-34}}{2.77 	imes 10^{-22}} \approx 2.39 	imes 10^{-11} \ m$.Converting to $\mathring{A}$s: $2.39 	imes 10^{-11} \ m = 0.239 \ \mathring{A} \approx 0.24 \ \mathring{A}$.

Similar Questions

What should be the order of arrangement of de-Broglie wavelength of an electron $(\lambda_{e})$,an $\alpha$-particle $(\lambda_{\alpha})$,and a proton $(\lambda_{p})$,given that all have the same kinetic energy?

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

Two particles of equal masses are moving with equal speeds at an angle $60^o$. The de-Broglie wavelength of these particles is $\lambda$. Find the de-Broglie wavelength of the particles in the frame of the centre of mass of the particles.

Two large parallel plates are connected to a $100 \, V$ power supply. These plates have a fine hole at the centre. An electron having energy $200 \, eV$ is directed such that it passes through the holes. When it emerges, its de-Broglie wavelength is ............. $\mathring{A}$

An alpha particle moves along a circular path of radius $0.5 \ mm$ in a magnetic field of $2 \times 10^{-2} \ T$. The de Broglie wavelength associated with the alpha particle is nearly (Planck's constant $= 6.63 \times 10^{-34} \ J \ s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module