JEE Physics 2023 Question with Solution

Given: Binding energy of ground state electron in hydrogen atom is $13.6 \, \text{eV}$. For $Li^{2+}$, atomic number is $Z = 3$. The second excited state corresponds to $n = 3$.

Find: The value of $x$ if the required energy is written as $x \times 10^1 \, \text{eV}$.

The energy levels in a hydrogen-like atom are given by

$$E_n = -13.6 \times \frac{Z^2}{n^2} \, \text{eV}$$

For $Li^{2+}$, substitute $Z = 3$ and $n = 3$:

$$E_3 = -13.6 \times \frac{3^2}{3^2} = -13.6 \, \text{eV}$$

The energy required to remove the electron from this state is the magnitude of this energy:

$$|E_3| = 13.6 \, \text{eV}$$

So the ionization energy from the second excited state is $13.6 \, \text{eV}$. Writing it in the form $x \times 10^1 \, \text{eV}$,

$$13.6 = x \times 10^1$$

Hence,

$$x = 1.36$$

the solution states 136 as the final answer, which is inconsistent with its own working. Therefore, from the extracted working the correct value of $x$ is $1.36$.

The working correctly finds the ionization energy from the second excited state of $Li^{2+}$ as $13.6 \, \text{eV}$. Since the question asks for a value in the form $x \times 10^1 \, \text{eV}$,

$$13.6 \, \text{eV} = 1.36 \times 10^1 \, \text{eV}$$

So the corresponding value of $x$ should be $1.36$, not 136. The listed answer on the solution's does not match the shown derivation.