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

$A$ particle of charge $q$ and mass $m$ enters a region of a transverse electric field of $E_{0} \hat{j}$ with initial velocity $v_{0} \hat{i}$. The time taken for the change in the de-Broglie wavelength of the charge from the initial value of $\lambda_{0}$ to $\lambda_{0} / 3$ is proportional to

$A$ long pipe is filled with small granules of nickel. The interatomic spacing for nickel is $d$. $A$ beam of electrons having different energies is incident on the left flat face of the pipe. Mass of electron is $m$. The beam of electrons coming out of the right flat face has energy:

If the momentum of an electron is changed by $\Delta P,$ then the de Broglie wavelength associated with it changes by $0.5\%.$ The initial momentum of the electron will be

If a change in the momentum of an electron by $P_o$ results in a change of $0.25\%$ in its de-Broglie wavelength,then the initial momentum is:

$A$ nucleus of mass $M$ at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2M'}{3}$ (where $M' < M$). The ratio of the de Broglie wavelengths of the two parts will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module