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Question

Out of a photon and an electron, the equation $$E = Pc$$, is valid for


A
both
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B
neither
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C
photon only
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D
electron only
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Solution

The correct option is C photon only
Energy is given by
$$E=\dfrac { { m }_{ 0 }{ c }^{ 2 } }{ \sqrt { 1-\dfrac { { v }^{ 2 } }{ { c }^{ 2 } }  }  } $$
$$\Rightarrow { E }^{ 2 }=\dfrac { { m }_{ 0 }^{ 2 }{ c }^{ 6 } }{ { c }^{ 2 }-{ v }^{ 2 } } $$
Momentum $$P$$ is given by
$$P=\dfrac { { m }_{ 0 }v }{ \sqrt { 1-\left( \dfrac { { v }^{ 2 } }{ { c }^{ 2 } }  \right)  }  } $$
$${ P }^{ 2 }{ c }^{ 2 }=\dfrac { { m }_{ 0 }^{ 2 }{ c }^{ 4 }{ v }^{ 2 } }{ { c }^{ 2 }-{ v }^{ 2 } } $$
$$\therefore { E }^{ 2 }-{ P }^{ 2 }{ c }^{ 2 }={ m }_{ 0 }^{ 2 }{ c }^{ 4 }$$
$$\Rightarrow { E }^{ 2 }={ P }^{ 2 }{ c }^{ 2 }+{ m }_{ 0 }^{ 2 }{ c }^{ 4 }$$
For photon, rest mass
$${ m }_{ 0 }=0$$, so $$E=Pc$$
For electron, $${ m }_{ 0 }\neq 0$$, so $$E\neq Pc$$

Physics
NCERT
Standard XII

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