
When the electron of a hydrogen atom jumps from ${{n = 4}}$ to ${{n = 1}}$ state, the number of spectral lines emitted will be:
A. $15$
B. $6$
C. $3$
D. $4$
Answer
559.8k+ views
Hint:Electrons get excited when it absorbs energy. When electrons are excited, it jumps from ground state to excited state. The ground state has a lower energy while the excited state has a higher energy. The energy is released when the electrons return back to the ground state.
Complete answer:
When the electron transitions occur in atoms or ions, spectral lines are formed. These spectral lines have different wavelengths.
There are six series by which electron transition occurs. They are Lymann, Balmer, Paschen, Brackett, Pfund and Humphrey series.
When electrons get excited, photons are emitted. The energy of photon is given below:
${{h}}\upsilon = {{{E}}_{{f}}} - {{{E}}_{{i}}}$, where ${{h}}$ is the Planck’s constant, $\upsilon $ is the frequency, ${{{E}}_{{f}}}$ is the final energy state and ${{{E}}_{{i}}}$ is the initial energy state.
It is given that the electron of a hydrogen atom jumps from ${{n = 4}}$ to ${{n = 1}}$ state.
Based on Bohr model of hydrogen spectrum, the number of spectral lines emitted can be calculated by the formula given below:
\[ = \dfrac{{\left( {{{{n}}_2} - {{{n}}_1}} \right)\left( {{{{n}}_2} - {{{n}}_1} - 1} \right)}}{2} = \dfrac{{\left( {4 - 1} \right)\left( {4 - 1 + 1} \right)}}{2} = \dfrac{{3 \times 4}}{2} = 6\]
When the electron of a hydrogen atom jumps from ${{n = 4}}$ to ${{n = 1}}$ state, the number of spectral lines will be $6$.
Hence, the correct option is B.
Note:
Different series of spectral lines have different uses. For example, the Balmer series has been used in the astronomy field since it appeared in stellar objects. Sometimes, in the question, ${{{n}}_1}$ may not be given. In this case, another formula can be used. The formula is $\dfrac{{{{n}}\left( {{{n}} - 1} \right)}}{2}$, where ${{n}}$ is the higher energy level. Sometimes, the electron comes back from excited state to ground state directly. Sometimes it may take several steps. But the energy released is the same as that the energy absorbed.
Complete answer:
When the electron transitions occur in atoms or ions, spectral lines are formed. These spectral lines have different wavelengths.
There are six series by which electron transition occurs. They are Lymann, Balmer, Paschen, Brackett, Pfund and Humphrey series.
When electrons get excited, photons are emitted. The energy of photon is given below:
${{h}}\upsilon = {{{E}}_{{f}}} - {{{E}}_{{i}}}$, where ${{h}}$ is the Planck’s constant, $\upsilon $ is the frequency, ${{{E}}_{{f}}}$ is the final energy state and ${{{E}}_{{i}}}$ is the initial energy state.
It is given that the electron of a hydrogen atom jumps from ${{n = 4}}$ to ${{n = 1}}$ state.
Based on Bohr model of hydrogen spectrum, the number of spectral lines emitted can be calculated by the formula given below:
\[ = \dfrac{{\left( {{{{n}}_2} - {{{n}}_1}} \right)\left( {{{{n}}_2} - {{{n}}_1} - 1} \right)}}{2} = \dfrac{{\left( {4 - 1} \right)\left( {4 - 1 + 1} \right)}}{2} = \dfrac{{3 \times 4}}{2} = 6\]
When the electron of a hydrogen atom jumps from ${{n = 4}}$ to ${{n = 1}}$ state, the number of spectral lines will be $6$.
Hence, the correct option is B.
Note:
Different series of spectral lines have different uses. For example, the Balmer series has been used in the astronomy field since it appeared in stellar objects. Sometimes, in the question, ${{{n}}_1}$ may not be given. In this case, another formula can be used. The formula is $\dfrac{{{{n}}\left( {{{n}} - 1} \right)}}{2}$, where ${{n}}$ is the higher energy level. Sometimes, the electron comes back from excited state to ground state directly. Sometimes it may take several steps. But the energy released is the same as that the energy absorbed.
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