De-broglie's wavelength of rest particle

Classical Mechanics Level 3

A particle 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

\dfrac{\sqrt{m_1}}{ \sqrt{m_2}}

\dfrac{M}{m_1 + m_2}

\dfrac{\sqrt{m_2}}{ \sqrt{m_1}}

1.0

\dfrac{m_1}{ m_2}

\dfrac{m_2}{ m_1}

None of these

2 solutions

Sahil Bansal
Jan 6, 2016

By the principle of conservation of momentum,

$P_{initial} = P_{final}$ $\Rightarrow M*0={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }\\ \Rightarrow { m }_{ 1 }{ v }_{ 1 }=-{ m }_{ 2 }{ v }_{ 2 }$

where ${ v }_{ 1 }$ and ${ v }_{ 2 }$ are the velocities of ${ m }_{ 1 }$ and ${ m }_{ 2 }$

So, $\left| { m }_{ 1 }{ v }_{ 1 } \right| =\left| { m }_{ 2 }{ v }_{ 2 } \right|$

Hence ${ \lambda }_{ 1 }={ \lambda }_{ 2 }\quad \because \quad \lambda =h/p$ as both masses have same momentum.

$\therefore \quad { \lambda }_{ 1 }/{ \lambda }_{ 2 }=1.$

Michael Mendrin
Jan 6, 2016

Momentum is conserved. Meanwhile, deBroglie wavelength is solely defined by momentum. Ergo, ratio is 1.0

De-broglie's wavelength of rest particle

2 solutions

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