The ground state energy of $H$ atom is $ - 13.6$ . The energy of second excited state $H{e^ + }$ in $eV$ is
A. $ - 6.04$
B. $ - 27.2$
C. $ - 54.4$
D. $ - 3.4$
Answer
597.9k+ views
Hint: Hydrogen is a single electron species. A hydrogen atom consists of a single positively charged proton and a single negatively charged electron. They are bound to the nucleus by the coulomb force.
If a neutral hydrogen atom loses its electron, it gets converted into a cation. The resulting ion, which consists solely of a proton for the usual isotope called hydron.
If instead a hydrogen atom gains a second electron, it get convert into an anion and called hydride
Complete step by step answer:
The formula used for the energy levels of a Hydrogen atom is:
$\left[ {\dfrac{{ - {E_0}{Z^2}}}{{{n^2}}}} \right]E = \left[ {\dfrac{{ - {E_0}{Z^2}}}{{{n^2}}}} \right]$
Where,
$
{E_0} = 13.6eV \\
1eV = 1.602 \times {10^{ - 19}}J \\
$
$n = 1,2,3......$
Energy of second excited state ( $H{e^ + }$) = For $H{e^ + }$ $n = 1,z = 2$
$\left[ {\dfrac{{ - {E_0}{Z^2}}}{{{n^2}}}} \right] = \dfrac{{ - 13.6 \times 4}}{1} \Rightarrow - 54.4eV$
Note:
The ground state means the lowest energy level $n = 1$ in which the atom is the most stable. The electron normally occupies this level if it does not have sufficient energy to move up to a higher level. Excited state means higher energy levels. When an atom gets excited from the ground state to a higher energy, it gets unstable and again falls back to one of the lower energy levels by the emission of photons that are electromagnetic radiations. The energy difference between any two adjacent levels gets smaller as $n$ increases, which results in the higher energy levels getting very close and crowded together just below $n = \infty $ . When an excited electron returns from higher level to a lower level, it releases an exact amount of energy by emitting a photon. The Lyman series of spectral lines corresponding to electron transitions from higher energy levels to level $n = 1$ Transitions to $n = 2$ and $n = 3$ are called the Balmer and Paschen series, respectively.
If a neutral hydrogen atom loses its electron, it gets converted into a cation. The resulting ion, which consists solely of a proton for the usual isotope called hydron.
If instead a hydrogen atom gains a second electron, it get convert into an anion and called hydride
Complete step by step answer:
The formula used for the energy levels of a Hydrogen atom is:
$\left[ {\dfrac{{ - {E_0}{Z^2}}}{{{n^2}}}} \right]E = \left[ {\dfrac{{ - {E_0}{Z^2}}}{{{n^2}}}} \right]$
Where,
$
{E_0} = 13.6eV \\
1eV = 1.602 \times {10^{ - 19}}J \\
$
$n = 1,2,3......$
Energy of second excited state ( $H{e^ + }$) = For $H{e^ + }$ $n = 1,z = 2$
$\left[ {\dfrac{{ - {E_0}{Z^2}}}{{{n^2}}}} \right] = \dfrac{{ - 13.6 \times 4}}{1} \Rightarrow - 54.4eV$
Note:
The ground state means the lowest energy level $n = 1$ in which the atom is the most stable. The electron normally occupies this level if it does not have sufficient energy to move up to a higher level. Excited state means higher energy levels. When an atom gets excited from the ground state to a higher energy, it gets unstable and again falls back to one of the lower energy levels by the emission of photons that are electromagnetic radiations. The energy difference between any two adjacent levels gets smaller as $n$ increases, which results in the higher energy levels getting very close and crowded together just below $n = \infty $ . When an excited electron returns from higher level to a lower level, it releases an exact amount of energy by emitting a photon. The Lyman series of spectral lines corresponding to electron transitions from higher energy levels to level $n = 1$ Transitions to $n = 2$ and $n = 3$ are called the Balmer and Paschen series, respectively.
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