
What electron transition in the \[H{e^ + }\] spectrum would have the same wavelength as the first Lymann transition of hydrogen?
Answer
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Hint:The Lyman Series consists of transitions that begin or end with the hydrogen ground state; the accompanying graphic depicts the atomic transitions that result in these two emission series. When an electron transitions from a higher energy state \[\left( {{n_h} = 2,3,4,5,6,...} \right)\] to a lower energy state \[\left( {{n_l} = 1} \right)\] , the Lyman series appears. The Lyman series' wavelengths are all in the Ultraviolet band.
Complete step by step answer:
Atomic electron transition is a change of state of an electron from one energy level to another within an atom. It appears to be discontinuous as the electron "jumps" from one energy level to another, it takes only a few nanoseconds to change its state from one energy level to another. \[H{e^ + }\] is an electron system with only one electron. A system's energy level can be written as
$E = R - h \times \dfrac{{{Z^2}}}{{{n^2}}},$
where \[E\] is the energy of a single atom.
\[R - h = 1.36\,ev\]
First Lyman series of hydrogen atom means
\[Z = \] atomic number, with \[n = 1\] for the \[H\] atom and similarly for helium atoms, we have \[z = 2\] for the \[H{e^ + }\] atom. \[n\] stands for the primary quantum number.
\[H\] has energy in the first Lyman transition: Let us calculate the energy level of the first Lyman transition of hydrogen as well as the helium atom.On solving for energy levels we have
$E\left( H \right) = R - h \times \dfrac{{1 - 1}}{{{2^2}}} \\
\therefore E = \left( {H{e^ + }} \right) = R - h \times {2^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right) \\ $
Since both sides of the equation are equal, the transition is indeed the first line in the Lyman series of the hydrogen spectrum. Hence, the transitions of the Lyman series in the hydrogen spectrum have the same wavelength as for helium for \[n = 4\] to \[n = 2\] , the electronic transition will take place, and the \[H{e^ + }\] spectrum will have the same wavelength as the first Lyman transition of \[H\] .
Additional information: The hydrogen atom is the most basic atomic system found in nature, it produces the most basic series. When a slit allows a beam of light or other radiation to enter the device, each component of the light or radiation forms an image of the source. When resolved under the spectroscope, these images can be seen. The photos will be in the shape of parallel lines with consistent spacing positioned next to each other. When moving from a higher to a lower wavelength side, the lines will be farther apart in the higher wavelength side and eventually closer. The shortest wavelength has the fewest separated spectral lines, which is referred to as the series limit.
Note: The hydrogen makes up the majority of the universe, its spectrum is very essential in astronomy. Series are sequences of lines corresponding to atomic transitions, each ending or beginning with the same atomic state in hydrogen, as a result of emission or absorption events in hydrogen.
Complete step by step answer:
Atomic electron transition is a change of state of an electron from one energy level to another within an atom. It appears to be discontinuous as the electron "jumps" from one energy level to another, it takes only a few nanoseconds to change its state from one energy level to another. \[H{e^ + }\] is an electron system with only one electron. A system's energy level can be written as
$E = R - h \times \dfrac{{{Z^2}}}{{{n^2}}},$
where \[E\] is the energy of a single atom.
\[R - h = 1.36\,ev\]
First Lyman series of hydrogen atom means
\[Z = \] atomic number, with \[n = 1\] for the \[H\] atom and similarly for helium atoms, we have \[z = 2\] for the \[H{e^ + }\] atom. \[n\] stands for the primary quantum number.
\[H\] has energy in the first Lyman transition: Let us calculate the energy level of the first Lyman transition of hydrogen as well as the helium atom.On solving for energy levels we have
$E\left( H \right) = R - h \times \dfrac{{1 - 1}}{{{2^2}}} \\
\therefore E = \left( {H{e^ + }} \right) = R - h \times {2^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right) \\ $
Since both sides of the equation are equal, the transition is indeed the first line in the Lyman series of the hydrogen spectrum. Hence, the transitions of the Lyman series in the hydrogen spectrum have the same wavelength as for helium for \[n = 4\] to \[n = 2\] , the electronic transition will take place, and the \[H{e^ + }\] spectrum will have the same wavelength as the first Lyman transition of \[H\] .
Additional information: The hydrogen atom is the most basic atomic system found in nature, it produces the most basic series. When a slit allows a beam of light or other radiation to enter the device, each component of the light or radiation forms an image of the source. When resolved under the spectroscope, these images can be seen. The photos will be in the shape of parallel lines with consistent spacing positioned next to each other. When moving from a higher to a lower wavelength side, the lines will be farther apart in the higher wavelength side and eventually closer. The shortest wavelength has the fewest separated spectral lines, which is referred to as the series limit.
Note: The hydrogen makes up the majority of the universe, its spectrum is very essential in astronomy. Series are sequences of lines corresponding to atomic transitions, each ending or beginning with the same atomic state in hydrogen, as a result of emission or absorption events in hydrogen.
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