Question

The series limit for the Balmer series of hydrogen spectrum occurs at $3664{\text{ }}{{\text{A}}^ \circ }$. Calculate
A. the ionization energy of hydrogen atom
B. the wavelength of the photon that would remove the electron in the ground state of the hydrogen atom.
(This question has multiple correct options)

$$(i){\text{ A}}{\text{. E = 5}}{\text{.42 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J}} \\\ {\text{ B}}{\text{. }}\lambda {\text{ = 916 }}{{\text{A}}^ \circ } \\\$$ $$(ii){\text{ A}}{\text{. E = 1}}{\text{.42 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J}} \\\ {\text{ B}}{\text{. }}\lambda {\text{ = 16 }}{{\text{A}}^ \circ } \\\$$ $$(iii){\text{ A}}{\text{. E = 5}}{\text{.42 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J}} \\\ {\text{ B}}{\text{. wave number = 27 }} \times {\text{ 1}}{{\text{0}}^5}{\text{ }}{{\text{m}}^{ - 1}} \\\$$ $$(iv){\text{ A}}{\text{. E = 3}}{\text{.38 eV}} \\\ {\text{ B}}{\text{. wave number = 109 }} \times {\text{ 1}}{{\text{0}}^5}{\text{ }}{{\text{m}}^{ - 1}} \\\$$

Answer 1

For the Balmer series the electron can make transitions from the second energy level. If an electron comes from a higher energy level then it will always jump to $n = 2$ always. For finding ionization energy of an atom the electron will fall from infinity to $n = 2$. The amount of ionisation energy will be equal to the energy of an electron in ground state for finding the wavelength of photon required to remove that electron from ground state.
Formula Used:
Energy of photon $= {\text{ }}\dfrac{{hc}}{\lambda }$

Complete answer: Let ${n_1}$ be the final energy level of an electron and ${n_2}$ be the energy level from which it falls. For obtaining a Balmer series of spectral lines the value of ${n_1}$ will always be equal to two. For finding the ionisation energy for an electron ${n_2}$ will be equal to infinity. Therefore it can be represented as,
${n_1} = 2$ , ${n_2} = {\text{ }}\infty$. The photon energy of the series can be find as,
$\Delta E{\text{ = }}{{\text{E}}_\infty }{\text{ - }}{{\text{E}}_2}{\text{ = - }}{{\text{E}}_2}$
$\Delta E{\text{ = - }}{{\text{E}}_2}$
$\Delta E{\text{ = - }}\dfrac{{hc}}{\lambda }$
Now, we know that $h{\text{ = 6}}{\text{.626 }} \times {\text{ 1}}{{\text{0}}^{ - 34}}{\text{ J s}}$, ${\text{c = 3 }} \times {\text{ 1}}{{\text{0}}^8}{\text{ m }}{{\text{s}}^{ - 1}}$ and wavelength is given which is equal to $3664{\text{ }}{{\text{A}}^ \circ }$. Substitute the values in the equation as,
$\Delta E{\text{ = - }}\dfrac{{6.626{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 34}}{\text{ }} \times {\text{ 3 }} \times {\text{ 1}}{{\text{0}}^8}}}{{3664{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 10}}}}$
$\Delta E{\text{ = - 5}}{\text{.42 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J}}$
Energy will be equal to $\Delta E{\text{ }} \times {\text{ }}{{\text{n}}^2}$. Therefore it can be calculated as
Energy ${\text{ = }}\Delta E{\text{ }} \times {\text{ }}{{\text{n}}^2}$
Energy ${\text{ = 5}}{\text{.42 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ }} \times {\text{ 4}}$
Energy ${\text{ = - 21}}{\text{.68 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J }}$
Thus we get the ionisation energy for a given photon. Now we will calculate the wavelength of the photon which is required to remove an electron from the ground state. This can be calculated as,
${\text{ - 21}}{\text{.68 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J = }}\dfrac{{hc}}{\lambda }$
On putting the values of h and c we get the result as,
${\text{ - 21}}{\text{.68 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J = }}\dfrac{{6.626{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 34}}{\text{ }} \times {\text{ 3 }} \times {\text{ 1}}{{\text{0}}^8}{\text{ }}}}{\lambda }$
$\lambda {\text{ = }}\dfrac{{6.626{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 34}}{\text{ }} \times {\text{ 3 }} \times {\text{ 1}}{{\text{0}}^8}{\text{ }}}}{{{\text{ - 21}}{\text{.68 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J}}}}$
$\lambda {\text{ = 916 }}{{\text{A}}^ \circ }$
The energy can be converted into eV by dividing it by $1.6{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 19}}$. So we get energy in electron-volts as
Energy ${\text{ = }}\dfrac{{{\text{ - 21}}{\text{.68 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J}}}}{{1.6{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 19}}}} = {\text{ 3}}{\text{.38 eV}}$
Also wave number is the reciprocal of the wavelength, therefore wave number can be calculated as
Wave number ${\text{ = }}\dfrac{1}{{916{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 10}}}} = {\text{ 109 }} \times {\text{ 1}}{{\text{0}}^5}{\text{ }}{{\text{m}}^{ - 1}}$
Thus the correct options are A. and D.

Note:
h is known as Planck constant and c is known as speed of light. These are constants which have fixed value. For finding energy in electron volts we divide the energy by electronic charge. Wave number is the reciprocal of wavelength, that’s why it's unit is ${m^{ - 1}}$.