By Sunil Bhardwaj

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From the maxwell distribution law, $$ \frac { { dn }_{ c } }{ n } =4\pi { \left( \frac { M }{ 2\pi RT } \right) }^{ \frac { 3 }{ 2 } }{ exp }^{ \left( \frac { -M{ c }^{ 2 } }{ 2RT } \right) }{ c }^{ 2 }dc $$ $$ \therefore \frac { { dn }_{ c } }{ n } =4\pi { \left( \frac { M }{ 2\pi RT } \right) }^{ \frac { 3 }{ 2 } }{ exp }^{ \left( \frac { -E }{ RT } \right) }{ c }^{ 2 }dc\qquad ......(1) $$ where \(\frac { 1 }{ 2 } M{ c }^{ 2 }=E\) or Kinetic Energy of gas molecules. $$ \therefore M{ c }^{ 2 }=2E\qquad ......(2) $$ Takein sq root of eq 2 $$ { M }^{ \frac { 1 }{ 2 } }c={ \left( 2E \right) }^{ \frac { 1 }{ 2 } }\qquad ......(3) $$ Also taking derivative of eq 2 $$ 2Mc dc = 2 dE $$ $$ Mc dc = dE\qquad ......(4) $$ Now multiplying eq 3 with 4 $$ { M }^{ \frac { 1 }{ 2 } }c \times Mc dc = { \left( 2E \right) }^{ \frac { 1 }{ 2 } } \times dE $$ $$ { M }^{ \frac { 3 }{ 2 } }{ c }^{ 2 } dc = { \left( 2E \right) }^{ \frac { 1 }{ 2 } } dE $$ $$ \therefore { c }^{ 2 } dc = \frac { { \left( 2E \right) }^{ \frac { 1 }{ 2 } } }{ { M }^{ \frac { 3 }{ 2 } } } dE \qquad ......(5) $$ Substituting ths value in eq 1, we get $$ \frac { { dn }_{ c } }{ n } =4\pi { \left( \frac { M }{ 2\pi RT } \right) }^{ \frac { 3 }{ 2 } }{ exp }^{ \left( \frac { -E }{ RT } \right) }\frac { { \left( 2E \right) }^{ \frac { 1 }{ 2 } } }{ { M }^{ \frac { 3 }{ 2 } } } dE $$ $$ \therefore \frac { 1 }{ n } \frac { { dn }_{ c } }{ dE } =4\pi \frac { { \left( M \right) }^{ \frac { 3 }{ 2 } } }{ { \left( 2\pi RT \right) }^{ \frac { 3 }{ 2 } } } { exp }^{ \left( \frac { -E }{ RT } \right) }\frac { { \left( 2E \right) }^{ \frac { 1 }{ 2 } } }{ { M }^{ \frac { 3 }{ 2 } } } $$ $$ \therefore \frac { 1 }{ n } \frac { { dn }_{ c } }{ dE } =\frac { { 4\pi } }{ 2\sqrt { 2 } { \left( \pi RT \right) }^{ \frac { 3 }{ 2 } } } { exp }^{ \left( \frac { -E }{ RT } \right) }{ \sqrt { 2 } \left( E \right) }^{ \frac { 1 }{ 2 } } $$ $$ \therefore \frac { 1 }{ n } \frac { { dn }_{ c } }{ dE } =\frac { { 2\pi } }{ { \left( \pi RT \right) }^{ \frac { 3 }{ 2 } } } { exp }^{ \left( \frac { -E }{ RT } \right) }{ \left( E \right) }^{ \frac { 1 }{ 2 } } $$ $$ \therefore p =\frac { { 2\pi } }{ { \left( \pi RT \right) }^{ \frac { 3 }{ 2 } } } { exp }^{ \left( \frac { -E }{ RT } \right) }{ \left( E \right) }^{ \frac { 1 }{ 2 } }\qquad ......(6) $$ Where \(p =\frac { 1 }{ n } \frac { { dn }_{ c } }{ dE } \) represents probability of molecules having energy E