Find the % difference (approximate) in de Bro | 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 $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.Since the kinetic energy of a gas molecule is $K.E. = \frac{3}{2}kT$,w

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.Since the kinetic energy of a gas molecule is $K.E. = \frac{3}{2}kT$,we have $\lambda \propto \frac{1}{\sqrt{T}}$.Let the initial temperature be $T_1 = T$ and the final temperature be $T_2 = 2T$.Then,$\lambda_1 = \frac{k'}{\sqrt{T}}$ and $\lambda_2 = \frac{k'}{\sqrt{2T}} = \frac{\lambda_1}{\sqrt{2}}$.The percentage change is given by $\frac{\lambda_2 - \lambda_1}{\lambda_1} 	imes 100 = \left( \frac{1}{\sqrt{2}} - 1 ight) 	imes 100$.Using $\frac{1}{\sqrt{2}} \approx 0.707$,the change is $(0.707 - 1) 	imes 100 = -29.3 \%$.The approximate magnitude of the percentage difference is $30 \%$.

Find the $\%$ difference (approximate) in de Broglie wavelength if the temperature of a gas becomes double.

Similar Questions

The kinetic energy (in $J$) of a particle of mass $4.5 \times 10^{-31} \ kg$ having a wavelength of $1000 \ nm$ is : $(h = 6.62 \times 10^{-34} \ J \ s)$

If the kinetic energy of an electron is $2.8 \times 10^{-23} \ J$,then the de Broglie wavelength will be ....... $(m_e = 9.1 \times 10^{-31} \ kg)$.

If the wavelength of a photon is $2.2 \times 10^{-11} \ m$ and $h = 6.6 \times 10^{-34} \ J \ s$,then the momentum of the photon is:

Given the ratio of kinetic energy of an electron in two orbitals is $16:9$. Calculate the ratio of the wavelength of the electron waves.

The wave nature of an electron was first given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module