Question 265: Atoms & Nuclei

Question

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 :

Accepted Answer

In hydrogen atom,
E
n
= $${{ - {E_0}} \over {{n^2}}}$$
Where E
0
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 = E
n+1
- E
n
= $${E_0}\left( {{1 \over {{n^2}}} + {1 \over {{{\left( {n + 1} ight)}^2}}}} ight)$$
= $${E_0}\left( {{{{{\left( {n + 1} ight)}^2} - {n^2}} \over {{n^2}{{\left( {n + 1} ight)}^2}}}} ight)$$
= E
0
$$\left( {{{2n + 1} \over {{n^4}{{\left( {1 + {1 \over n}} ight)}^2}}}} ight)$$
= E
0
$$\left( {{{n\left( {2 + {1 \over n}} ight)} \over {{n^4}{{\left( {1 + {1 \over n}} ight)}^2}}}} ight)$$
Since n >>> 1
Hence, $${{1 \over n} \simeq 0}$$
= $${E_0}\left( {{2 \over {{n^3}}}} ight)$$
As $$\Delta $$E = h$$
u $$
$$ 	herefore $$ h$$
u $$ = $${E_0}\left( {{2 \over {{n^3}}}} ight)$$
$$ \Rightarrow $$ $$
u $$ $$ \propto $$ $${1 \over {{n^3}}}$$

Answer

$${1 \over n}$$

Answer

$${1 \over {{n^2}}}$$

Answer

$${1 \over {{n^3}}}$$

Answer

$${1 \over {{n^4}}}$$

Atoms & Nuclei

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 :

Choose the correct answer: