Chemistry Q72 – JEE-MAIN 2025

The energy of an electron in first Bohr orbit of H-atom is

- 13.6 eV

. The magnitude of energy value of electron in the first excited state of

{Be}^{3 +}

is _______

eV

(nearest integer value)

Enter Calculated Numerical Value:

Recall the Bohr model for hydrogen-like species, which describes the energy levels of electrons in atoms or ions with only one electron.

Step 1: Identify the relevant formula and parameters✦ Active

The energy of an electron in the $n$ -th orbit of a hydrogen-like species is given by the formula:

E_{n} = - 13.6 \frac{Z^{2}}{n^{2}} eV

For the ${Be}^{3 +}$ ion:

- The atomic number $Z$ for Beryllium (Be) is 4.

- The first excited state corresponds to the principal quantum number $n = 2$ (ground state is $n = 1$ ).

Step 2: Calculate the energy value○ Expand

Substitute the values of $Z = 4$ and $n = 2$ into the energy formula:

E_{2} = - 13.6 \frac{4^{2}}{2^{2}}

E_{2} = - 13.6 \frac{16}{4}

E_{2} = - 13.6 \times 4

E_{2} = - 54.4 eV

Step 3: Determine the magnitude and round to the nearest integer○ Expand

The question asks for the magnitude of the energy value. The magnitude is the absolute value of the calculated energy:

| E_{2} | = | - 54.4 eV | = 54.4 eV

Rounding to the nearest integer, the magnitude of the energy is $54 eV$ .

💡 Teacher's Secret Hint

Remember to take the magnitude as requested by the question and round to the nearest integer.

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Chemistry Question 72 – JEE-MAIN 2025