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

An electron beam has a kinetic energy equal to $100\; eV$. Find its wavelength (in $\mathring{A}$) associated with the beam,given the mass of an electron $= 9.1 \times 10^{-31}\; kg$,$1\; eV = 1.6 \times 10^{-19}\; J$,and Planck's constant $= 6.6 \times 10^{-34}\; Js$.

How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$? (Let $E$ be the initial energy of the electron).

If a proton and an electron have the same linear momentum,then compared to the electron:

The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^8\, m/s$ is equal to the wavelength of a photon. The ratio of the kinetic energy of the particle to the energy of the photon is (velocity of light is $3 \times 10^8\, m/s$). (in $/8$)

If the velocity of an electron increases,what will be the change in its de Broglie wavelength?

Vedclass Test Series

Exam Paper Generator

Online Exam Module