
Ionization potential of hydrogen atom is $13.6\,eV$ . Hydrogen atoms in the ground state are excited by the monochromatic radiation of photon energy $12.1\,eV$ . According to the Bohr’s theory, the spectral lines emitted by hydrogen will be:
A.Two
B.Three
C.Four
D.One
Answer
559.8k+ views
Hint:
Ionization energy is the minimum amount of energy required to remove an electron from an isolated atom. Principal quantum numbers are the numbers greater than zero. Principal quantum numbers are also called as the shells that are assigned to each electron in an atom.
Formula used:
$E=\dfrac{-13.6}{{{n}^{2}}}$
where, $E$ is the energy of an electron and $n$ is the principal quantum number.
Number of spectral lines $=\,\dfrac{n\left( n-1 \right)}{2}$
where, $n$ is the principal quantum number.
Complete step by step answer:
Ionisation potential of hydrogen atom is $13.6\,eV$
The excited photon energy provided is $12.1\,eV$
As we know that ionization energy is equal to $-13.6\,eV$
Energy of an electron in the excited state = ionization energy + photon energy
On substituting the values in the above formula, we get, energy of an electron in the excited state $=-13.6+12.1$
$\Rightarrow -1.5\,eV$
Also, the energy of an electron in ${{n}^{th}}$ orbit is:
$E=\dfrac{-13.6}{{{n}^{2}}}$
where, $E$ is the energy of electron and $n$ is the principal quantum number.
On substituting the values in the above formula, we get,
$-1.5=\dfrac{-13.6}{{{n}^{2}}}$
On further solving, we get
${{n}^{2}}=\dfrac{-13.6}{-1.5}$
$\Rightarrow {{n}^{2}}=9$
$\Rightarrow n=3$
Therefore, the principal quantum number $(n)=3$
To calculate spectral lines, the formula used,
Number of spectral lines = $\dfrac{n\left( n-1 \right)}{2}$
On substituting the values, we get number of spectral lines $=\dfrac{3(3-1)}{2}$
On further solving, we get number of spectral lines $=3$
Therefore, the correct option is (B), that is, three.
Additional information:
-Hydrogen atom is the simplest atom that has one proton and one electron. When an electron moves from one energy level to another energy level, then the energy of the electron also changes. Energy is required by the electron to move from one state to another.
-Ionization energy is defined as the amount of energy required to remove an electron from an isolated atom or molecule. Ionization energy is expressed in Joules or electron volt $(eV)$
Note: Principal quantum number is a quantum number that is assigned to each electron in an atom that describes the electron state. It is denoted with a symbol, $n$ . The ionization energy of a hydrogen atom is $13.6\,eV$ . Factors affecting the ionization energy are nuclear charge, number of electron shells, electron configuration.
Ionization energy is the minimum amount of energy required to remove an electron from an isolated atom. Principal quantum numbers are the numbers greater than zero. Principal quantum numbers are also called as the shells that are assigned to each electron in an atom.
Formula used:
$E=\dfrac{-13.6}{{{n}^{2}}}$
where, $E$ is the energy of an electron and $n$ is the principal quantum number.
Number of spectral lines $=\,\dfrac{n\left( n-1 \right)}{2}$
where, $n$ is the principal quantum number.
Complete step by step answer:
Ionisation potential of hydrogen atom is $13.6\,eV$
The excited photon energy provided is $12.1\,eV$
As we know that ionization energy is equal to $-13.6\,eV$
Energy of an electron in the excited state = ionization energy + photon energy
On substituting the values in the above formula, we get, energy of an electron in the excited state $=-13.6+12.1$
$\Rightarrow -1.5\,eV$
Also, the energy of an electron in ${{n}^{th}}$ orbit is:
$E=\dfrac{-13.6}{{{n}^{2}}}$
where, $E$ is the energy of electron and $n$ is the principal quantum number.
On substituting the values in the above formula, we get,
$-1.5=\dfrac{-13.6}{{{n}^{2}}}$
On further solving, we get
${{n}^{2}}=\dfrac{-13.6}{-1.5}$
$\Rightarrow {{n}^{2}}=9$
$\Rightarrow n=3$
Therefore, the principal quantum number $(n)=3$
To calculate spectral lines, the formula used,
Number of spectral lines = $\dfrac{n\left( n-1 \right)}{2}$
On substituting the values, we get number of spectral lines $=\dfrac{3(3-1)}{2}$
On further solving, we get number of spectral lines $=3$
Therefore, the correct option is (B), that is, three.
Additional information:
-Hydrogen atom is the simplest atom that has one proton and one electron. When an electron moves from one energy level to another energy level, then the energy of the electron also changes. Energy is required by the electron to move from one state to another.
-Ionization energy is defined as the amount of energy required to remove an electron from an isolated atom or molecule. Ionization energy is expressed in Joules or electron volt $(eV)$
Note: Principal quantum number is a quantum number that is assigned to each electron in an atom that describes the electron state. It is denoted with a symbol, $n$ . The ionization energy of a hydrogen atom is $13.6\,eV$ . Factors affecting the ionization energy are nuclear charge, number of electron shells, electron configuration.
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