Question

Energy of an electron is given by E = \[ - 2.178{\text{ }} \times {\text{ }}{10^{ - 18\;}}J\dfrac{{{Z^2}}}{{{n^2}}}.\] Wavelength of light required to excite an electron in a hydrogen atom from level n=1 to n=2 will be:
[h =\[\;6.62{\text{ }} \times {10^{ - 34}}\]Js and c \[ = {\text{ }}3.0 \times {10^8}\]\[m{s^{ - 1}}\]]
\[\begin{gathered}
\left( A \right)\;\;2.816 \times {10^{ - 7}}m \\
\left( B \right)\;\;6.500 \times {10^{ - 7}}m \\
\left( C \right)\;\;8.500 \times {10^{ - 7}}m \\
\left( D \right)\;\;1.214 \times {10^{ - 7}}m \\
\end{gathered} \]\[\]

Answer 1

Hint: The energy required to excite an electron from one energy state to another is provided by the light energy of suitable wavelength. Hence the energy difference between the first and second level of the hydrogen atom is equal to the energy of light.

Complete step by step solution:
Step-1: First of all find the given data from the question:
Energy of electron (E)= \[ - 2.178{\text{ }} \times {\text{ }}{10^{ - 18\;}}J\dfrac{{{Z^2}}}{{{n^2}}}.\]
h =\[\;6.62{\text{ }} \times {10^{ - 34}}\]Js
c \[ = {\text{ }}3.0 \times {10^8}\]\[m{s^{ - 1}}\]
$\lambda $ (Wavelength) of light required to excite an electron in a hydrogen atom from level 1 to 2 =?
Step-2
 Now calculate the energy of the first and second energy level of the hydrogen atom.
-Energy of electron in first shell: \[{E_1}\]= \[ - 2.178{\text{ }} \times {\text{ }}{10^{ - 18\;}}J\dfrac{{{Z^2}}}{{{n^2}}}.\]
=\[\; - 2.178{\text{ }} \times {\text{ }}{10^{ - 18}}\dfrac{{{1^2}}}{{{1^2}}}J\]
= \[ - 2.178{\text{ }} \times {\text{ }}{10^{ - 18\;}}J\]

-Energy of electron in second shell: \[{E_2}\]= \[ - 2.178{\text{ }} \times {\text{ }}{10^{ - 18}}\]\[\dfrac{{{Z^2}}}{{{n^2}}}\]J
= \[ - 2.178{\text{ }} \times {\text{ }}{10^{ - 18}}^{}\]$\dfrac{{{1^2}}}{{{2^2}}}$J
= \[ - 2.178{\text{ }} \times {\text{ }}{10^{ - 18}}^{}\]×$\dfrac{1}{4}$J
= \[ - 0.5{\text{ }} \times {\text{ }}{10^{ - 18}}J\]
Step-3:
 Now, calculate energy difference between the shells
\[\;\Delta E = {\text{ }}{E_2}-{\text{ }}{E_1}\]
= (\[ - 0.5{\text{ }} \times {\text{ }}{10^{ - 18}}J\]- )-(\[ - 2.178{\text{ }} \times {\text{ }}{10^{ - 18\;}}J\])
= \[1.6335 \times {\text{ }}{10^{ - 18}}J\]
Step-4:
This energy difference is equal to the energy of light.
Hence, $\Delta E = E$
E = $\dfrac{{hc}}{\lambda }$
\[1.6335 \times {\text{ }}{10^{ - 18}}J\]= $\dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda }$
$\lambda $ = \[1.214 \times {10^{ - 7}}m\]

So, the correct option is (D) i.e. \[1.214 \times {10^{ - 7}}m\]

Note: When an electron excites, it absorbs energy of a particular wavelength. When this electron comes down back to the ground state from the excited one, it emits energy of the same wavelength which one it absorbed earlier. Energy of the light of this wavelength is equal to the energy difference of the energy levels between those the electron is moving.