What is the energy required to move the electron from the ground states of the H atom to the first excited state? Given that the ground states energy of H atom is $ - 13.6\,eV$ and that the energy ${E_n}$ of electron in ${n^{th}}$ orbital of an atom or an ion of atomic number Z is, given by the equation ${E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}$
A. $13.6\,eV$
B. $3.4\,eV$
C. $10.2\,eV$
D. $ - 10.2\,eV$
Answer
535.8k+ views
Hint:In order to solve this question we need to understand that electron distribution in atom was first proposed by Niels Bohr, according to him the angular momentum of electron is integral multiple of $\dfrac{h}{{2\pi }}$ and from here he calculated the radius of electron and energy of electron, it was found that electron distributed in various shells around the nucleus. So in this question we will find the energy of the required level and then calculate the difference between two levels.
Complete step by step answer:
Atomic number of hydrogen is, $Z = 1$. So Energy of ${n^{th}}$ orbital of a hydrogen atom is, given by the equation
${E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}$
Putting value of Z we get, ${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}$
For ground state energy, $n = 1$ so ground state energy is given by,
${E_1} = - 13.6\,eV$
For first excited state energy, $n = 2$ so first excited energy is given as,
${E_2} = \dfrac{{ - 13.6}}{{{{(2)}^2}}}\,eV$
$\Rightarrow {E_2} = \dfrac{{ - 13.6}}{4}\,eV$
$\Rightarrow {E_2} = - 3.2eV$
So the energy required to move the electron from ground to first orbit is,
$E = {E_2} - {E_1}$
Putting values we get,
$E = - 3.2eV - ( - 13.6eV)$
$\Rightarrow E = - 3.2eV + 13.6eV$
$\therefore E = 10.4\,eV$
So the correct option is C.
Note: It should be remembered that we have calculated the difference because to excite the electron to the next level we have to supply the energy difference between two shells. Electron is excited by striking a photon having energy equal to the difference between two levels. When an electron is excited to a higher energy level then it de-excites and falls back to previous level.
Complete step by step answer:
Atomic number of hydrogen is, $Z = 1$. So Energy of ${n^{th}}$ orbital of a hydrogen atom is, given by the equation
${E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}$
Putting value of Z we get, ${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}$
For ground state energy, $n = 1$ so ground state energy is given by,
${E_1} = - 13.6\,eV$
For first excited state energy, $n = 2$ so first excited energy is given as,
${E_2} = \dfrac{{ - 13.6}}{{{{(2)}^2}}}\,eV$
$\Rightarrow {E_2} = \dfrac{{ - 13.6}}{4}\,eV$
$\Rightarrow {E_2} = - 3.2eV$
So the energy required to move the electron from ground to first orbit is,
$E = {E_2} - {E_1}$
Putting values we get,
$E = - 3.2eV - ( - 13.6eV)$
$\Rightarrow E = - 3.2eV + 13.6eV$
$\therefore E = 10.4\,eV$
So the correct option is C.
Note: It should be remembered that we have calculated the difference because to excite the electron to the next level we have to supply the energy difference between two shells. Electron is excited by striking a photon having energy equal to the difference between two levels. When an electron is excited to a higher energy level then it de-excites and falls back to previous level.
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