A Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be.
A. 2
B. 3
C. 5
D. 6
Answer
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Hint: We have a hydrogen atom which is excited from its ground state. The principal quantum number of the excited atom is given to us. We are asked to find the number of spectral lines observed due to the excitation. We have an equation to find the number of spectral lines. By substituting the required values in the equation, we will get the total number of spectral lines.
Formula used:
\[N=\dfrac{n\left( n-1 \right)}{2}\]
Complete step by step answer:
In the question it is said that a hydrogen atom is excited from the ground state.
The principal quantum number is given to us as 4.
We are asked to find the number of spectral lines when the atom is excited.
We know that to find the total number of spectral lines we have an equation,
\[N=\dfrac{n\left( n-1 \right)}{2}\], where ‘N’ is the total number of spectral lines and ‘n’ is the principal quantum number.
From the question we know that $n=4$.
By substituting the value of ‘n’ in the equation, we will get
$\Rightarrow N=\dfrac{4\left( 4-1 \right)}{2}$
$\Rightarrow N=\dfrac{16-4}{2}$
$\Rightarrow N=\dfrac{12}{2}$
$\therefore N=6$
Therefore we get the total number of spectral lines when the hydrogen atom gets excited as 6.
Hence the correct answer is option D.
Note:
Quantum numbers are a set of numbers which we use to describe the position and energy of an electron in an atom. The four different quantum numbers are; principal quantum number, azimuthal quantum number, magnetic quantum number and spin quantum number. The principal quantum number describes the energy and the most probable distance of the electron from the nucleus. The azimuthal quantum number gives the shape of the orbital and therefore the angular distribution. Magnetic quantum number determines the total number of orbitals in the subshell and the orientation of these orbitals. The spin quantum number is independent of all the other quantum numbers and it gives us the direction of spin of the electron.
Formula used:
\[N=\dfrac{n\left( n-1 \right)}{2}\]
Complete step by step answer:
In the question it is said that a hydrogen atom is excited from the ground state.
The principal quantum number is given to us as 4.
We are asked to find the number of spectral lines when the atom is excited.
We know that to find the total number of spectral lines we have an equation,
\[N=\dfrac{n\left( n-1 \right)}{2}\], where ‘N’ is the total number of spectral lines and ‘n’ is the principal quantum number.
From the question we know that $n=4$.
By substituting the value of ‘n’ in the equation, we will get
$\Rightarrow N=\dfrac{4\left( 4-1 \right)}{2}$
$\Rightarrow N=\dfrac{16-4}{2}$
$\Rightarrow N=\dfrac{12}{2}$
$\therefore N=6$
Therefore we get the total number of spectral lines when the hydrogen atom gets excited as 6.
Hence the correct answer is option D.
Note:
Quantum numbers are a set of numbers which we use to describe the position and energy of an electron in an atom. The four different quantum numbers are; principal quantum number, azimuthal quantum number, magnetic quantum number and spin quantum number. The principal quantum number describes the energy and the most probable distance of the electron from the nucleus. The azimuthal quantum number gives the shape of the orbital and therefore the angular distribution. Magnetic quantum number determines the total number of orbitals in the subshell and the orientation of these orbitals. The spin quantum number is independent of all the other quantum numbers and it gives us the direction of spin of the electron.
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