GREPhysics.NET: GR8677 #68

GR8677 #68

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Special Relativity $\Rightarrow$ }Relativistic Energy

The relativistic energy relation is,

$E= \sqrt{p^2 c^2 + mc^4.}$

As the speed of a particle $v \rightarrow c$ , $E\rightarrow pc$ , since by the de Broglie relation, $p=h/\lambda=h\nu/c=E/c$ . Plug in the requirement into the energy relation to find that $m\rightarrow 0$ , as in choice (A).

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Comments

spacemanERAU
2009-10-20 08:37:41

Also, this problem is really simple if you remember that any particle that has speed of c MUST have rest mass of zero...no math involved

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radicaltyro
2006-10-23 00:41:07

It should be $E=\sqrt{p^2c^2+m^2c^4}$

spacemanERAU
2009-10-18 18:35:28

agreed!

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LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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