A material particle with rest mass ${m_ \circ }$is moving with a velocity of light $c$. Then the wavelength of the de-Broglie wave associated with it is-
A) $\dfrac{h}{{{m_ \circ }c}}$
B) Zero
C) $\infty $
D) $\dfrac{{{m_ \circ }c}}{h}$
Answer
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Hint: The de-Broglie wavelength of a particle is given by $\lambda = \dfrac{h}{{mv}}$ , where h=Planck’s constant, v= velocity and m=relativistic mass of the particle. If a body will moves with a speed comparable to that of speed of light , then the relativistic mass is given by-
$m = \dfrac{{{m_ \circ }}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}$. First calculate this mass and then calculate the de-Broglie wavelength.
Complete step by step solution:
Step1: As given that speed of the particle is c i.e. speed of light and from the theory of relativity if a body moves with a speed comparable to that of the speed of light then its mass is not constant and it is given by the following equation-
$m = \dfrac{{{m_ \circ }}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}$ …………….(1)
Where m=relativistic mass of the particle, ${m_ \circ }$=rest mass of the particle,
v=velocity of the particle and c=speed of light.
Since v=c (given in question), Therefore from eq.(1)
$
m = \dfrac{{{m_ \circ }}}{{\sqrt {1 - \dfrac{{{c^2}}}{{{c^2}}}} }} \\
m = \dfrac{{{m_ \circ }}}{{\sqrt {1 - 1} }} \\
m = \dfrac{{{m_ \circ }}}{{\sqrt 0 }} \\
m = \dfrac{{{m_ \circ }}}{0} \\
\Rightarrow m = \infty \\
$
It means the relativistic mass tends to zero.
Step2: Now use the de-Broglie wavelength equation to calculate the wavelength
Since de-Broglie wavelength is given by
$\lambda = \dfrac{h}{{mv}}$
Substituting all the values in the above equation, we have
$
\lambda = \dfrac{h}{{(\infty )c}} \\
\Rightarrow \lambda = 0 \\
$
Hence de-Broglie wavelength is zero.
$\therefore $ The correct option is (B).
Note:
Always remember that if a body is moving with a velocity comparable to that of the velocity of light then, according to the theory of relativity, mass is not constant and this theory of relativity is given by Albert Einstein. The mass which is taken in the formula of de Broglie wavelength is having the moving mass of the particle. Do not take the rest mass of the particle in the formula in place of m. Also, remember that the infinity in the denominator gives the result as zero.
$m = \dfrac{{{m_ \circ }}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}$. First calculate this mass and then calculate the de-Broglie wavelength.
Complete step by step solution:
Step1: As given that speed of the particle is c i.e. speed of light and from the theory of relativity if a body moves with a speed comparable to that of the speed of light then its mass is not constant and it is given by the following equation-
$m = \dfrac{{{m_ \circ }}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}$ …………….(1)
Where m=relativistic mass of the particle, ${m_ \circ }$=rest mass of the particle,
v=velocity of the particle and c=speed of light.
Since v=c (given in question), Therefore from eq.(1)
$
m = \dfrac{{{m_ \circ }}}{{\sqrt {1 - \dfrac{{{c^2}}}{{{c^2}}}} }} \\
m = \dfrac{{{m_ \circ }}}{{\sqrt {1 - 1} }} \\
m = \dfrac{{{m_ \circ }}}{{\sqrt 0 }} \\
m = \dfrac{{{m_ \circ }}}{0} \\
\Rightarrow m = \infty \\
$
It means the relativistic mass tends to zero.
Step2: Now use the de-Broglie wavelength equation to calculate the wavelength
Since de-Broglie wavelength is given by
$\lambda = \dfrac{h}{{mv}}$
Substituting all the values in the above equation, we have
$
\lambda = \dfrac{h}{{(\infty )c}} \\
\Rightarrow \lambda = 0 \\
$
Hence de-Broglie wavelength is zero.
$\therefore $ The correct option is (B).
Note:
Always remember that if a body is moving with a velocity comparable to that of the velocity of light then, according to the theory of relativity, mass is not constant and this theory of relativity is given by Albert Einstein. The mass which is taken in the formula of de Broglie wavelength is having the moving mass of the particle. Do not take the rest mass of the particle in the formula in place of m. Also, remember that the infinity in the denominator gives the result as zero.
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