A material particle with a rest mass m_0 is m | vedclass

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

Question

(B) The relativistic mass $m$ of a particle moving with velocity $v$ is given by $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{C^2}}}$.As the particle moves wi

Vedclass · Accepted Answer

Zero

Vedclass · Answer

(B) The relativistic mass $m$ of a particle moving with velocity $v$ is given by $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{C^2}}}$.As the particle moves with the velocity of light,$v = C$.Substituting this into the mass formula: $m = \frac{m_0}{\sqrt{1 - \frac{C^2}{C^2}}} = \frac{m_0}{\sqrt{1 - 1}} = \frac{m_0}{0} = \infty$.The de$-$Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Substituting the values,$\lambda = \frac{h}{\infty \cdot C} = 0$.Therefore,the de$-$Broglie wavelength associated with the particle is $0$.

$A$ material particle with a rest mass $m_0$ is moving with the velocity of light $C$. Then,the wavelength of the de$-$Broglie wave associated with it is

Similar Questions

The electron microscope is based on the principle of

The de-Broglie wavelength is proportional to

The ratio of wavelengths of a proton and a deuteron accelerated by potentials $V_{p}$ and $V_{d}$ is $1 : \sqrt{2}$. Then,the ratio of $V_{p}$ to $V_{d}$ will be:

The de-Broglie wavelength of an electron accelerated by a potential of $50\,V$ is approximately ............. $\mathring{A}$. $(|e| = 1.6 \times 10^{-19}\,C, m_e = 9.1 \times 10^{-31}\,kg, h = 6.6 \times 10^{-34}\,Js)$

An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{e} / \lambda_{p}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module