Question
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 and m2 with non-zero velocities. The ratio λ1/λ2 of de Broglie wavelengths of particles is
A. m2/m1
B. m1/m2
C. m2/m1
D. 1.0
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=λhc
Here, h is Planck’s constant, c is the speed of light and λ is the de Broglie wavelength.
The linear momentum of the particle of mass m moving with velocity v is,
p=mv
Complete step by step answer:
We know that, according to de-Broglie’s hypothesis, the momentum of the particle is,
p=λhc
Here, h is Planck’s constant, c is the speed of light and λ 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+m2v2
Here, v is the velocity of parent particle, v1 is the velocity m1 and m1 is the velocity of m2. Since the parent particle is at rest, the initial velocity v is zero. Therefore, the above equation becomes,
0=m1v1+m2v2
⇒m1v1=−m2v2
Therefore, from the above equation, the momentum of the particle of mass m1 and the momentum of the particle of mass m2 is equal.
So, we can write,
p1=p2
⇒λ1hc=λ2hc
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,
∴λ2λ1=1
So, the correct answer is option (D).
Note: The negative sign for v2 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.