Question

A partical of mass M at rest decays into two Particles of masses m_1 and m_2 having non-zero velocities. The ratio of the de - Broglie wavelengths of the particles lambda_1| lambda_2 is (a) m_1//m_2 (b)m_2// m_1 (c ) 1 (d)sqrt,_2 // sqrt_1

Answer 1

Answer:

b. Will be correct answer

Explanation:

wavelength is inversely proportional to momentum.

Answer 2

The ratio of the de - Broglie wavelengths = 1

Given:

A partical of mass M at rest decays into two Particles of masses $m_1$ and $m_2$ having non-zero velocities.

To find:

The ratio of the de - Broglie wavelengths of the particles.

Formula used:

De - Broglie wavelength = $\frac{h}{m_1 v_1}$

Conservation of linear momentum:

Initial momentum =  Final momentum

Explanation:

Mass M is in rest so initial velocity of mass = 0

So initial momentum = M×v = 0

Mass M decays into two Particles of masses $m_1$ and $m_2$

So  Final momentum ,= $m_{1} v_{1}-m_{2} v_{2}$

From conservation of linear momentum:

Initial momentum =  Final momentum

                         0 = $m_{1} v_{1}-m_{2} v_{2}$

                       $m_{1} v_{1}=m_{2} v_{2}$

De - Broglie wavelength = $\frac{h}{m_1 v_1}$

           $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{\hbar}{m_{1} v_{1}}}{\frac{h}{m_{2} v_{2}}}$

Here h is constant and $m_{1} v_{1}=m_{2} v_{2}$

So  The ratio of the de - Broglie wavelengths = 1

To learn more...

1)Two particles have equal momenta. What is the ratio of their de-broglie wavelengths?

https://brainly.in/question/6843886

2)Orbits of a particle moving in a circle are such that perimeter of the orbit equals an integer number of de-broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to:

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