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Question: A particle of mass M at rest decays into two masses \[{m_1}\] and \[{m_2}\] with non-zero velocities...

A particle of mass M at rest decays into two masses m1{m_1} and m2{m_2} with non-zero velocities. The ratio λ1/λ2{\lambda _1}/{\lambda _2} of de Broglie wavelengths of particles is
A. m2/m1{m_2}/{m_1}
B. m1/m2{m_1}/{m_2}
C. m2/m1\sqrt {{m_2}} /\sqrt {{m_1}}
D. 1.01.0

Explanation

Solution

According to de-Broglie’s hypothesis, the momentum of the particle is inversely proportional to the wavelength of the particle. Use the law of conservation of momentum to relate the momentum of both particles using de-Broglie’s hypothesis.

Formula used:
p=hcλp = \dfrac{{hc}}{\lambda }
Here, h is Planck’s constant, c is the speed of light and λ\lambda is the de Broglie wavelength.
The linear momentum of the particle of mass m moving with velocity v is,
p=mvp = mv

Complete step by step answer:
We know that, according to de-Broglie’s hypothesis, the momentum of the particle is,
p=hcλp = \dfrac{{hc}}{\lambda }
Here, h is Planck’s constant, c is the speed of light and λ\lambda is the de Broglie wavelength.

According to the law of conservation of momentum, the momentum of a system remains conserved.
Therefore, we can write,
Mv=m1v1+m2v2Mv = {m_1}{v_1} + {m_2}{v_2}
Here, v is the velocity of parent particle, v1{v_1} is the velocity m1{m_1} and m1{m_1} is the velocity of m2{m_2}. Since the parent particle is at rest, the initial velocity v is zero. Therefore, the above equation becomes,
0=m1v1+m2v20 = {m_1}{v_1} + {m_2}{v_2}
m1v1=m2v2\Rightarrow {m_1}{v_1} = - {m_2}{v_2}
Therefore, from the above equation, the momentum of the particle of mass m1{m_1} and the momentum of the particle of mass m2{m_2} is equal.

So, we can write,
p1=p2{p_1} = {p_2}
hcλ1=hcλ2\Rightarrow \dfrac{{hc}}{{{\lambda _1}}} = \dfrac{{hc}}{{{\lambda _2}}}
Planck’s constant h and speed of light c is constant for both particles. Therefore, the wavelength of these particles is the same. Therefore, we can write,
λ1λ2=1\therefore \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = 1

So, the correct answer is option (D).

Note: The negative sign for v2{v_2} implies that the motion of the second particle is opposite to the direction of the first particle. While solving these types of questions, students can blindly use the law of conservation of momentum. Since the Planck’s constant and speed of light is constant, you can directly take the ratio of wavelength by de-Broglie hypothesis.