
The Ionization potential of a hydrogen atom is \[13.6{\text{ eV}}\]. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy \[12.1{\text{ eV}}\]. According to Bohr's theory, the spectral lines emitted by hydrogen will be:
A.Two
B.Three
C.Four
D.One
Answer
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Hint: The ground state energy of the photon can be calculated by subtracting ionization potential with the energy of the photon in excited state. Using the formula we will calculate the number of spectral lines.
Formula used: \[{{\text{E}}_{\text{n}}} = - \dfrac{{13.6{\text{ }}{{\text{Z}}^2}}}{{{{\text{n}}^2}}}\]
Here \[{{\text{E}}_{\text{n}}}\] is the energy of a photon, Z for hydrogen is 1 and n will give us the value of spectral lines.
Complete step-by-step answer: A monochromatic light is that which consists of a single wavelength.
The electron only reaches the excited state when it has enough energy to cover the gap between excited state and the ground state. To overcome the gap we need to provide external energy equals to the ionization energy to reach excited state and hence,
The ionization potential is given to us as \[13.6{\text{ eV}}\] and the excited energy is \[12.1{\text{ eV}}\].
Hence with the given values in question we can calculate the ground state energy as
$ {\text{Ground state }} = (12.1 - 13.6){\text{ eV}} = - 1.5{\text{ eV}} $
Atomic number of hydrogen is 1. We will substitutes all the values in the formula as follow:
\[ - 1.5 = - \dfrac{{13.6{\text{ }}{{\text{1}}^2}}}{{{{\text{n}}^2}}}\]
Solving and rearranging we will get,
\[{{\text{n}}^2} = \dfrac{{13.6}}{{1.5}}\]
Taking square root both sides we will get,
\[{\text{n}} = 3\]
Hence the correct option is B.
Note: According to the Bohr model, electrons revolve in their respective stationary orbit, and their energy remains constant. Each orbit has a constant energy or quantized energy. When external energy is provided to the electron, the force of attraction between electron and nucleus loosen and the electron jumps to a higher energy level. The energy should be equal to the energy gap between the two energy levels. The term ionization means removing the electron.
Formula used: \[{{\text{E}}_{\text{n}}} = - \dfrac{{13.6{\text{ }}{{\text{Z}}^2}}}{{{{\text{n}}^2}}}\]
Here \[{{\text{E}}_{\text{n}}}\] is the energy of a photon, Z for hydrogen is 1 and n will give us the value of spectral lines.
Complete step-by-step answer: A monochromatic light is that which consists of a single wavelength.
The electron only reaches the excited state when it has enough energy to cover the gap between excited state and the ground state. To overcome the gap we need to provide external energy equals to the ionization energy to reach excited state and hence,
The ionization potential is given to us as \[13.6{\text{ eV}}\] and the excited energy is \[12.1{\text{ eV}}\].
Hence with the given values in question we can calculate the ground state energy as
$ {\text{Ground state }} = (12.1 - 13.6){\text{ eV}} = - 1.5{\text{ eV}} $
Atomic number of hydrogen is 1. We will substitutes all the values in the formula as follow:
\[ - 1.5 = - \dfrac{{13.6{\text{ }}{{\text{1}}^2}}}{{{{\text{n}}^2}}}\]
Solving and rearranging we will get,
\[{{\text{n}}^2} = \dfrac{{13.6}}{{1.5}}\]
Taking square root both sides we will get,
\[{\text{n}} = 3\]
Hence the correct option is B.
Note: According to the Bohr model, electrons revolve in their respective stationary orbit, and their energy remains constant. Each orbit has a constant energy or quantized energy. When external energy is provided to the electron, the force of attraction between electron and nucleus loosen and the electron jumps to a higher energy level. The energy should be equal to the energy gap between the two energy levels. The term ionization means removing the electron.
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