A praticle of mass M at rest decays into two particle of masses `m_1` and `m_2`, having non-zero velocities. The ratio of the de Broglie wavelength of the particles `(lamda_1)/(lamda_2)` is

Options

A. `m_(1)//m_(2)`
B. `m_(2)//m_(1)`
C. `1.0`
D. `sqrt(m_(2))//sqrt(m_(1))`

Answer

3

Explanation

By law of conservation of momentum
`0=m_(1)overset(to)(v_(1))+m_(2)overset(to)(v_(2))rArrm_(1)overset(to)(v_(1))=-m_(2)overset(to)(v_(2))`
-ve sign incidicates that both he particles are moving in opposite direction. Now de-Broglie wavelenght
`lambda_(1)=(h)/(m_(1)v_(1))andlambda_(2)=(h)/(m_(2)v_(2)),therefore(lambda_1)/(lambda_(2))=(m_(2)v_(2))/(m_(1)v_(1))=1`