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Electrons in a certain energy level $n = {n_1}$ ​, can emit 3 spectral lines. When they are at another energy level, $n = {n_2}$. They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio ofA. 4:3B. 3:4C. 2:1D. 1:2

Last updated date: 07th Sep 2024
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Answer
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Hint:The electron jumps to the lower energy level by radiating out energy in the form of spectral lines. The number of spectral lines is proportional to the possible number of transitions made by the electron to reach the ground state.

Formula used:
Using Bohr’s postulate the speed of electron in nth state is given as,
${v_n} = \dfrac{{2\pi KZ{e^2}}}{{nh}}$
The number of emission spectral lines emitted by the electron during transition from nth state to the ground state is given as,
$N = \dfrac{{n\left( {n - 1} \right)}}{2}$

Complete step by step solution:
When an electron is in nth state and jumps to the ground state then it releases energy in the form of radiation and emits the spectral lines. The number of emission spectral lines emitted by the electron during transition from nth state to the ground state is given as,
$N = \dfrac{{n\left( {n - 1} \right)}}{2}$
It is given that when the electron is in the state $n = {n_1}$ then it emits a total of 3 spectral lines.

Putting in the expression for the number of spectral lines emitted, we get
$3 = \dfrac{{{n_1}\left( {{n_1} - 1} \right)}}{2}$
On solving the equation, we get
$n_1^2 - {n_1} - 6 = 0$
$\Rightarrow n_1^2 - 3{n_1} + 2{n_1} - 6 = 0$
$\Rightarrow \left( {{n_1} - 3} \right)\left( {{n_1} + 2} \right) = 0$
As the state of the electron is a positive whole number, so the value of ${n_1}$ is 3.

Similarly for the state ${n_2}$ the number of spectral lines emitted is 6. Putting in the expression for the number of spectral lines emitted, we get
$6 = \dfrac{{{n_2}\left( {{n_2} - 1} \right)}}{2}$
On solving the equation, we get
$n_2^2 - {n_2} - 12 = 0$
$\Rightarrow n_2^2 - 4{n_2} + 3{n_2} - 12 = 0$
$\Rightarrow \left( {{n_2} - 4} \right)\left( {{n_1} + 3} \right) = 0$
Hence, the value of ${n_2}$ is 4.

Using Bohr’s postulate the speed of electron in nth state is given as,
${v_n} = \dfrac{{2\pi KZ{e^2}}}{{nh}}$
So, the ratio of the speeds of the electron in both the state is,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\dfrac{{2\pi KZ{e^2}}}{{{n_1}h}}}}{{\dfrac{{2\pi KZ{e^2}}}{{{n_2}h}}}} \\$
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{n_2}}}{{{n_1}}} \\$
$\therefore \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{4}{3}$
Hence, the ratio of the speed of the electron in both the states is 4:3.

Therefore, the correct option is A.

Note: We must be careful while choosing the solution of the quadratic equation. As the principal quantum number is a positive whole number, we need to choose only the positive solution of the quadratic equation.