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(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

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Zero

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(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

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