The de Broglie wavelength of an oxygen molecu | vedclass

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

Question

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.For a gas molecule, the average kinetic energy is $K = \frac{

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.For a gas molecule, the average kinetic energy is $K = \frac{3}{2}kT$.Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2m(\frac{3}{2}kT)}} = \frac{h}{\sqrt{3mkT}}$.Given values: $h = 6.63 	imes 10^{-34} \ J \cdot s$, $m = 5.31 	imes 10^{-26} \ kg$, $k = 1.38 	imes 10^{-23} \ J/K$, and $T = 27 + 273 = 300 \ K$.Calculating the denominator: $\sqrt{3 	imes 5.31 	imes 10^{-26} 	imes 1.38 	imes 10^{-23} 	imes 300} = \sqrt{658.734 	imes 10^{-49}} = \sqrt{65.8734 	imes 10^{-48}} \approx 8.116 	imes 10^{-24}$.$\lambda = \frac{6.63 	imes 10^{-34}}{8.116 	imes 10^{-24}} \approx 0.8168 	imes 10^{-10} \ m = 81.68 	imes 10^{-12} \ m$.Wait, re-calculating: $\sqrt{3 	imes 5.31 	imes 1.38 	imes 300} 	imes 10^{-24.5} \approx \sqrt{6587.34} 	imes 10^{-24.5} \approx 81.16 	imes 10^{-24.5}$.Correct calculation: $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{3 	imes 5.31 	imes 10^{-26} 	imes 1.38 	imes 10^{-23} 	imes 300}} = \frac{6.63 	imes 10^{-34}}{\sqrt{6587.34 	imes 10^{-49}}} = \frac{6.63 	imes 10^{-34}}{2.566 	imes 10^{-23}} \approx 2.58 	imes 10^{-11} \ m = 25.8 	imes 10^{-12} \ m$.Rounding to the nearest integer, $x = 26$.

The de Broglie wavelength of an oxygen molecule at $27^{\circ} C$ is $x \times 10^{-12} \ m$. The value of $x$ is (take Planck's constant $= 6.63 \times 10^{-34} \ J \cdot s$, Boltzmann constant $= 1.38 \times 10^{-23} \ J/K$, mass of oxygen molecule $= 5.31 \times 10^{-26} \ kg$).

Similar Questions

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

An electron is moving through a field. It is moving $(i)$ opposite to an electric field and $(ii)$ perpendicular to a magnetic field as shown. For each situation, determine the de-Broglie wavelength of the electron:

An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

$A$ wave is associated with matter:

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module