Question 664: Atoms & Nuclei

Question

The radius of the orbit of an electron in a hydrogen-like atom is 4.5a
0
, where a
0
is the Bohr radius. Its orbital angular momentum is ${{3h} \over {2\pi }}$. It is given that h is Planck constant and R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is(are)

Accepted Answer

The orbital angular momentum is
$L = {{nh} \over {2\pi }}$
${{3h} \over {2\pi }} = {{nh} \over {2\pi }} \Rightarrow n = 3$
The radius of the orbit is
$4.5({a_0}) = {a_0}\left( {{{{n^2}} \over Z}} ight)$
$Z = {{{n^2}} \over {4.5}} = {{{3^2}} \over {4.5}} = {9 \over {4.5}} = 2$
Thus, the possible transitions are 3$	o$2, 3$	o$1 and 2$	o$1. For the transition 3$	o$2, the wavelength is
${1 \over \lambda } = R{(2)^2}\left[ {{1 \over 4} - {1 \over 9}} ight] = 4R\left[ {{{9 - 4} \over {36}}} ight] = {{5R} \over 9}$
$ \Rightarrow \lambda  = {9 \over {5R}}$
For the transition 3$	o$1, the wavelength is
${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 9}} ight] = 4R\left( {{8 \over 9}} ight) = {{32R} \over 9}$
$ \Rightarrow \lambda  = {9 \over {32R}}$
For the transition 2$	o$1, the wavelength is
${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 4}} ight] = 4R\left( {{3 \over 4}} ight) = 3R$
$ \Rightarrow \lambda  = {1 \over {3R}}$

Answer

${9 \over {32R}}$

Answer

${9 \over {16R}}$

Answer

${9 \over {5R}}$

Answer

${4 \over {3R}}$

Atoms & Nuclei

The radius of the orbit of an electron in a hydrogen-like atom is 4.5a 0 , where a 0 is the Bohr radius. Its orbital angular momentum is {{3h} \over {2\pi }}. It is given that h is Planck constant and R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is(are)

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