JEE Chemistry 2025 Question with Solution

Given: The energy of an electron in first Bohr orbit of H-atom is $-13.6 \, \text{eV}$. For hydrogen-like species, the total energy is

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

Find: The magnitude of energy of electron in the first excited state of Be$^{3+}$.

For Be$^{3+}$, $Z = 4$ and the first excited state means $n = 2$.

Substitute these values:

$$E_2 = -13.6 \frac{4^2}{2^2} \, \text{eV}$$ $$E_2 = -13.6 \times \frac{16}{4} \, \text{eV}$$ $$E_2 = -13.6 \times 4 = -54.4 \, \text{eV}$$

The magnitude is

$$|E_2| = 54.4 \, \text{eV}$$

Nearest integer value is $54$. Therefore, the answer is $54$.

Given: For hydrogen-like ions, $E_n \propto \dfrac{Z^2}{n^2}$. For H-atom in the first orbit, $Z=1$ and $n=1$, so energy is $-13.6 \, \text{eV}$.

Find: The magnitude of energy in the first excited state of Be$^{3+}$.

For Be$^{3+}$, $Z=4$ and first excited state means $n=2$. Hence,

$$\frac{E_{\text{Be}^{3+}}}{E_H} = \frac{\left(\dfrac{4^2}{2^2}\right)}{\left(\dfrac{1^2}{1^2}\right)} = \frac{16/4}{1} = 4$$

So,

$$E_{\text{Be}^{3+}} = 4(-13.6) = -54.4 \, \text{eV}$$

Therefore, the magnitude of the energy is $54.4 \, \text{eV}$, which gives $54$ as the nearest integer.