At temperature T,the average kinetic energy o | vedclass

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

Question

(D) The kinetic energy of a particle is given by $K.E. = \frac{3}{2} KT = \frac{1}{2} mv^2$. Thus,$v = \sqrt{\frac{3KT}{m}}$. The de Broglie wavelengt

Vedclass · Accepted Answer

Visible photon $>$ Thermal electron $>$ Thermal neutron

Vedclass · Answer

(D) The kinetic energy of a particle is given by $K.E. = \frac{3}{2} KT = \frac{1}{2} mv^2$. Thus,$v = \sqrt{\frac{3KT}{m}}$. The de Broglie wavelength is $\lambda = \frac{h}{mv} = \frac{h}{m \sqrt{\frac{3KT}{m}}} = \frac{h}{\sqrt{3KTm}}$. This implies $\lambda \propto \frac{1}{\sqrt{m}}$. Since the mass of a photon is effectively zero (or it travels at $c$),it has the largest wavelength. Comparing particles,$m_{	ext{electron}} Therefore,$\lambda_{	ext{electron}} > \lambda_{	ext{neutron}} > \lambda_{	ext{proton}}$. Considering the photon,the order is $\lambda_{	ext{photon}} > \lambda_{	ext{electron}} > \lambda_{	ext{neutron}}$.

At temperature $T$,the average kinetic energy of any particle is $\frac{3}{2} KT$. The de Broglie wavelength follows the order:

Similar Questions

Calculate the $\lambda$ of $CO_2$ molecule moving with a velocity $440 \ m/s.$

The velocity associated with a proton moving in a potential difference of $1000 \, V$ is $4.37 \times 10^{5} \, ms^{-1}$. If a hockey ball of mass $0.1 \, kg$ is moving with this velocity,calculate the wavelength associated with this velocity.

What is the mass of a particle with a wavelength of $3.313 \mathring{A}$ moving with a speed of $2.0 \times 10^8 \ m \ s^{-1}$?

If the radius of the first orbit of hydrogen atom is $a_0$, then de Broglie's wavelength of electron in $3^{rd}$ orbit is (in $\pi a_0$)

For the quantization of the Bohr orbit,its circumference will be equal to .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module