JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 14)
In a hydrogen atom the electron makes a
transition from (n + 1)th level to the nth level.
If n >> 1, the frequency of radiation emitted is
proportional to :
$${1 \over n}$$
$${1 \over {{n^2}}}$$
$${1 \over {{n^3}}}$$
$${1 \over {{n^4}}}$$
Explanation
In hydrogen atom,
En = $${{ - {E_0}} \over {{n^2}}}$$
Where E0 is Ionisation Energy of H.
For transition from (n + 1) to n, the energy of emitted radiation is equal to the difference in energies of levels.
$$\Delta $$E = En+1 - En
= $${E_0}\left( {{1 \over {{n^2}}} + {1 \over {{{\left( {n + 1} \right)}^2}}}} \right)$$
= $${E_0}\left( {{{{{\left( {n + 1} \right)}^2} - {n^2}} \over {{n^2}{{\left( {n + 1} \right)}^2}}}} \right)$$
= E0$$\left( {{{2n + 1} \over {{n^4}{{\left( {1 + {1 \over n}} \right)}^2}}}} \right)$$
= E0$$\left( {{{n\left( {2 + {1 \over n}} \right)} \over {{n^4}{{\left( {1 + {1 \over n}} \right)}^2}}}} \right)$$
Since n >>> 1
Hence, $${{1 \over n} \simeq 0}$$
= $${E_0}\left( {{2 \over {{n^3}}}} \right)$$
As $$\Delta $$E = h$$\nu $$
$$ \therefore $$ h$$\nu $$ = $${E_0}\left( {{2 \over {{n^3}}}} \right)$$
$$ \Rightarrow $$ $$\nu $$ $$ \propto $$ $${1 \over {{n^3}}}$$
En = $${{ - {E_0}} \over {{n^2}}}$$
Where E0 is Ionisation Energy of H.
For transition from (n + 1) to n, the energy of emitted radiation is equal to the difference in energies of levels.
$$\Delta $$E = En+1 - En
= $${E_0}\left( {{1 \over {{n^2}}} + {1 \over {{{\left( {n + 1} \right)}^2}}}} \right)$$
= $${E_0}\left( {{{{{\left( {n + 1} \right)}^2} - {n^2}} \over {{n^2}{{\left( {n + 1} \right)}^2}}}} \right)$$
= E0$$\left( {{{2n + 1} \over {{n^4}{{\left( {1 + {1 \over n}} \right)}^2}}}} \right)$$
= E0$$\left( {{{n\left( {2 + {1 \over n}} \right)} \over {{n^4}{{\left( {1 + {1 \over n}} \right)}^2}}}} \right)$$
Since n >>> 1
Hence, $${{1 \over n} \simeq 0}$$
= $${E_0}\left( {{2 \over {{n^3}}}} \right)$$
As $$\Delta $$E = h$$\nu $$
$$ \therefore $$ h$$\nu $$ = $${E_0}\left( {{2 \over {{n^3}}}} \right)$$
$$ \Rightarrow $$ $$\nu $$ $$ \propto $$ $${1 \over {{n^3}}}$$
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