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  • A particle of mass at rest decays into two particles of masses <img alt="\mathrm{m_1 \: and\: m_2}" s

A particle of mass \mathrm{M} at rest decays into two particles of masses \mathrm{m_1 \: and\: m_2} having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles \mathrm{\lambda_1 / \lambda_2} is :

 

Option: 1

\mathrm{\frac{m_1}{m_2}}
 


Option: 2

\mathrm{\frac{m_2}{m_1}}
 


Option: 3

1


Option: 4

\mathrm{\frac{\sqrt{m_2}}{\sqrt{m_1}}}


Answers (1)

best_answer

From the law of conservation of momentum, \mathrm{P_1=P_2} (in opposite directions)
Now de-Broglie wavelength is given by

\mathrm{\lambda=\frac{h}{P} } where h= Planck's constant

Since, momentum \mathrm{(P)} of both the particles is equal, therefore

\mathrm{ \lambda_1 =\lambda_2 }

\mathrm{ \text { or } \lambda_1 / \lambda_2 =1 }

 

Posted by

Kuldeep Maurya

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