A material particle with a rest mass m0 is moving with a velocity of light c. Then, the wavelength of the de Broglie wave associated with it is

Question

A material particle with a rest mass m0 is moving with a velocity of light c. Then, the wavelength of the de Broglie wave associated with it is (A) (h / m0 c) (B) zero (C) ∞ (D) (m0 c / h)

Tardigrade Admin · Accepted Answer

We know that mass m in motion and the rest mass m0 is related through the equation m=(m0/√1-(v2)c2) As v=c, m=(m0/√1-1)=(m0/0)=∞ Therefore, de Broglie wavelength is λ=(h/m v)=(h/(∞)(c))=0

Q. A material particle with a rest mass $m_{0}$ is moving with a velocity of light $c$ . Then, the wavelength of the de Broglie wave associated with it is

$(h / m_{0} c)$

zero

$\infty$

$(m_{0} c / h)$

We know that mass $m$ in motion and the rest mass $m_{0}$ is related through the equation
$m = \frac{m _{0}}{1 - \frac{v ^{2}}{c ^{2}}}$
As $v = c, m = \frac{m _{0}}{1 - 1} = \frac{m _{0}}{0} = \infty$
Therefore, de Broglie wavelength is
$λ = \frac{h}{m v} = \frac{h}{( \infty ) ( c )} = 0$

Q. A material particle with a rest mass m0​ is moving with a velocity of light c. Then, the wavelength of the de Broglie wave associated with it is

(h/m0​c)