JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 5)
The number density of molecules of a gas depends on their distance r from the origin as, $$n\left( r \right) = {n_0}{e^{ - \alpha {r^4}}}$$.
Then the total number of molecules is proportional to :
$${n_0}{\alpha ^{ - 3/4}}$$
$${n_0}{\alpha ^{ - 3}}$$
$${n_0}{\alpha ^{1/4}}$$
$$\sqrt {{n_0}} {\alpha ^{1/2}}$$
Explanation
Lets take an element hollow sphere of thickness dr
Vol. of element dV = 4$$\pi $$r2dr
Total number of molecules,
N = $$\int\limits_0^\infty {n\,dV} $$
= $$\int\limits_0^\infty {{n_0}{e^{ - \alpha {r^4}}}\,4\pi {r^2}dr} $$
Let $${{e^{ - \alpha {r^4}}}}$$ = t ......................(1)
$$ \therefore $$ $$ - 4\alpha {e^{ - \alpha {r^4}}}{r^3}dr$$ = dt
$$ \Rightarrow $$ $${{dt} \over { - 4\alpha r}} = {e^{ - \alpha {r^4}}}{r^2}dr$$ .....................(2)
Taking $$\ln $$ to the both sides of the equation (1), we get
$${\ln}\left( t \right) = - \alpha {r^4}$$
$$ \Rightarrow $$ $$r = {\left( {{{\ln t} \over { - \alpha }}} \right)^{{1 \over 4}}}$$ .......(3)
Putting this value of r in equation (2),
$${{dt} \over { - 4\alpha {{\left( {{{\ln t} \over { - \alpha }}} \right)}^{{1 \over 4}}}}} = {e^{ - \alpha {r^4}}}{r^2}dr$$
$$ \Rightarrow $$ $${{{\alpha ^{{{ - 3} \over 4}}}dt} \over { - 4{{\left( { - \ln t} \right)}^{{1 \over 4}}}}} = {e^{ - \alpha {r^4}}}{r^2}dr$$
Putting in the integration, we get
N = $${n_0}4\pi \int\limits_1^0 {{{{\alpha ^{{{ - 3} \over 4}}}dt} \over { - 4{{\left( { - \ln t} \right)}^{{1 \over 4}}}}}} $$
= $$ - {n_0}{\alpha ^{{{ - 3} \over 4}}}\pi \int\limits_1^0 {{{dt} \over {{{\left( { - \ln t} \right)}^{{1 \over 4}}}}}} $$
$$ \therefore $$ N $$ \propto $$ $${n_0}{\alpha ^{{{ - 3} \over 4}}}$$
Vol. of element dV = 4$$\pi $$r2dr
Total number of molecules,
N = $$\int\limits_0^\infty {n\,dV} $$
= $$\int\limits_0^\infty {{n_0}{e^{ - \alpha {r^4}}}\,4\pi {r^2}dr} $$
Let $${{e^{ - \alpha {r^4}}}}$$ = t ......................(1)
$$ \therefore $$ $$ - 4\alpha {e^{ - \alpha {r^4}}}{r^3}dr$$ = dt
$$ \Rightarrow $$ $${{dt} \over { - 4\alpha r}} = {e^{ - \alpha {r^4}}}{r^2}dr$$ .....................(2)
Taking $$\ln $$ to the both sides of the equation (1), we get
$${\ln}\left( t \right) = - \alpha {r^4}$$
$$ \Rightarrow $$ $$r = {\left( {{{\ln t} \over { - \alpha }}} \right)^{{1 \over 4}}}$$ .......(3)
Putting this value of r in equation (2),
$${{dt} \over { - 4\alpha {{\left( {{{\ln t} \over { - \alpha }}} \right)}^{{1 \over 4}}}}} = {e^{ - \alpha {r^4}}}{r^2}dr$$
$$ \Rightarrow $$ $${{{\alpha ^{{{ - 3} \over 4}}}dt} \over { - 4{{\left( { - \ln t} \right)}^{{1 \over 4}}}}} = {e^{ - \alpha {r^4}}}{r^2}dr$$
Putting in the integration, we get
N = $${n_0}4\pi \int\limits_1^0 {{{{\alpha ^{{{ - 3} \over 4}}}dt} \over { - 4{{\left( { - \ln t} \right)}^{{1 \over 4}}}}}} $$
= $$ - {n_0}{\alpha ^{{{ - 3} \over 4}}}\pi \int\limits_1^0 {{{dt} \over {{{\left( { - \ln t} \right)}^{{1 \over 4}}}}}} $$
$$ \therefore $$ N $$ \propto $$ $${n_0}{\alpha ^{{{ - 3} \over 4}}}$$
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