Question

Energy of a stationary electron in the hydrogen atom is
\[ E=-\frac{13.6}{n^2}\ \text{eV} \] then the energies required to excite the electron in hydrogen atom to (a) its second excited state and (b) ionized state respectively are

• (a) \(\sim10\ \text{eV}\), (b) \(13.6\ \text{eV}\)
• (a) \(\sim12\ \text{eV}\), (b) \(13.6\ \text{eV}\)
• (a) \(\sim12\ \text{eV}\), (b) \(10.6\ \text{eV}\)
• (a) \(\sim8\ \text{eV}\), (b) \(13.6\ \text{eV}\)

Hint:

In the hydrogen atom, ground state corresponds to \(n=1\), first excited state to \(n=2\), and second excited state to \(n=3\).

Answer 1

The Correct Option is B

Step 1: Write the energy formula for hydrogen atom.
The energy of the electron in the \(n^{\text{th}}\) orbit is
\[ E_n=-\frac{13.6}{n^2}\ \text{eV} \]
Ground state corresponds to
\[ n=1 \]
So, ground state energy is
\[ E_1=-13.6\ \text{eV} \]

Step 2: Find the energy of the second excited state.
Second excited state corresponds to
\[ n=3 \]
Thus,
\[ E_3=-\frac{13.6}{3^2} \]
\[ E_3=-\frac{13.6}{9} \]
\[ E_3\approx-1.51\ \text{eV} \]
Energy required to excite the electron from ground state to second excited state is
\[ \Delta E=E_3-E_1 \]
\[ \Delta E=(-1.51)-(-13.6) \]
\[ \Delta E=12.09\ \text{eV} \]
\[ \Delta E\approx12\ \text{eV} \]

Step 3: Find the ionization energy.
For ionization, electron is taken to
\[ n=\infty \]
At infinity,
\[ E_\infty=0 \]
Therefore, ionization energy is
\[ \Delta E=0-(-13.6) \]
\[ \Delta E=13.6\ \text{eV} \]

Step 4: Final conclusion.
Hence, the required energies are
\[ \boxed{(a)\ \sim12\ \text{eV},\quad (b)\ 13.6\ \text{eV}} \]