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1st Equation of Motion

Assuming the ...

Question

Assuming the table contains no major errors, what can we conclude about the $(m a s s \times c^{2})$ of the undetected third particle?

A

It is

0.79 M e V .

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B

It is

0.39 M e V .

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C

It is less than or equal to

0.79 M e V

; but we cannot be more precise.

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D

It is less than or equal to

0.39 M e V

; but we cannot be more precise.

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Solution

The correct option is D It is less than or equal to $0.39 M e V$ ; but we cannot be more precise.
Total energy of the particle is $940.97 - (939.67 + 0.51 + 0.39 + 0.01) = 0.39 M e V$
which is both $K . E$ and $m a s s \times c^{2}$
Hence, $m a s s \times c^{2} \leq 0.39 M e V$

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Q. When subatomic particles undergo reactions, energy is conserved, but mass is not necessarily conserved. However, a particle’s mass “contributes” to its total energy, in accordance with Einstein’s famous equations,

E = m c^{2}

In this equation, E denotes the energy a particle carries because of its mass. The particle can also have additional energy due to its motion and its interactions with other particles.
Consider a neutron at rest, and well separated from other particles. It decays into a proton, an electron and an undetected third particle:
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Assuming table 1 contains no major errors, what can we conclude about the mass

\times c^{2}

of the undetected third particle?

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