Question

An electron revolves around a nucleus of charge \[ + Ze\]. If the energy required to excite the electron from the second to third Bohr orbit is \[47.2\,{\text{eV}}\], then the energy required to excite the electron from \[n = 3\] to \[n = 4\] state will be:
(A) \[16.53\,{\text{eV}}\]
(B) \[13.6\,{\text{eV}}\]
(C) \[1.51\,{\text{eV}}\]
(D) None of the above

Answer 1

Hint: First, we will find the expression of energy change in going from second to third orbit and simplify a bit. We will again find the energy change in going from third to fourth and use the previous equation in this present equation. We will manipulate and simplify accordingly.

Complete step by step answer:
In the given question,
The charge of the nucleus around which an electron revolves is \[ + Ze\] .
The energy required to excite the electron from the second to the third orbit is \[47.2\,{\text{eV}}\] .
We are asked to find the energy required to excite the electron from third to fourth energy level (i.e. state).
For this we will apply the formula which gives energy of an electron in a particular orbit, which is given by:
$E = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}$ …… (1)
Where,
\[Z\] indicates the atomic number of the element.
\[n\] indicates the energy level.
The energy found in the above expression is in electron volts.
Since, we are given that the energy required to excite the electron from the second to the third orbit is \[47.2\,{\text{eV}}\] .
Mathematically we can write:
\[{E_3} - {E_2} = 47.2\,{\text{eV}}\] …… (2)
Where,
\[{E_3}\] indicates the energy at the third level.
\[{E_2}\] indicates the energy at the second level.
Equation (2) be rewritten as:
${E_3} - {E_2} = 47.2\,{\text{eV}} \\$
$\Rightarrow\dfrac{{ - 13.6{Z^2}}}{{{3^2}}} - \dfrac{{ - 13.6{Z^2}}}{{{2^2}}} = 47.2 \\$
$\Rightarrow\dfrac{{ - 13.6{Z^2}}}{{{3^2}}} + \dfrac{{13.6{Z^2}}}{{{2^2}}} = 47.2 \\$
$\Rightarrow 13.6{Z^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) = 47.2 \\ $
Manipulating the above expression further, we get:
$13.6{Z^2}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) = 47.2 \\
\Rightarrow 13.6{Z^2} \times \dfrac{5}{{36}} = 47.2 \\$
\[\Rightarrow 13.6{Z^2} = 47.2 \times \dfrac{{36}}{5}\] …… (3)

Now, we will find the energy required to excite the electron from third state to fourth state:
$
\Rightarrow{E_4} - {E_3}= \dfrac{{ - 13.6{Z^2}}}{{{4^2}}} - \dfrac{{ - 13.6{Z^2}}}{{{3^2}}} \\
\Rightarrow{E_4} - {E_3}= \dfrac{{13.6{Z^2}}}{{{3^2}}} - \dfrac{{13.6{Z^2}}}{{{4^2}}} \\
\Rightarrow{E_4} - {E_3}= 13.6{Z^2}\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right) \\
$
Simplifying further we get:
$
\Rightarrow{E_4} - {E_3}= 13.6{Z^2}\left( {\dfrac{1}{9} - \dfrac{1}{{16}}} \right) \\
\Rightarrow{E_4} - {E_3}= 13.6{Z^2} \times \dfrac{7}{{144}} \\
\Rightarrow{E_4} - {E_3}= 47.2 \times \dfrac{{36}}{5} \times \dfrac{7}{{144}} \\
\therefore{E_4} - {E_3}= 16.52\,{\text{eV}} \\
$
Hence, the energy required to excite the electron from third state to fourth state is $16.52\,{\text{eV}}$. The correct option is (A).

Note:While solving the problem, do not worry about the atomic number of the element. As you proceed inside the solution, this physical quantity will be replaced by other terms. It is important to note that an electron absorbs energy to move from lower orbit to higher orbit while it loses energy to move from higher to lower. Negative sign indicates energy lost by it and positive sign indicates energy absorbed by it.