Chemistry Q52 – NEET-UG 2024

Home Chemistry Structure of AtomQuestion #607

The energy of an electron in the ground state (

n = 1

) for

{He}^{+}

ion is

- x

J, then that for an electron in

n = 2

state for

{Be}^{3 +}

ion in J is :

Clue 1 of 3: Bohr's Model for Hydrogen-like Species

The energy of an electron in a single-electron species (hydrogen-like) depends on the atomic number ( $Z$ ) and the principal quantum number ( $n$ ).

🧠 Full Solution Path

Step 1: Relate given energy to Rydberg constant✦ Active

For ${He}^{+}$ ion, the atomic number $Z = 2$ and the principal quantum number $n = 1$ (ground state). The energy is given as $- x$ J. Using the formula for the energy of an electron in a hydrogen-like species, $E_{n} = - R_{H} \frac{Z^{2}}{n^{2}}$ :

- x = - R_{H} \frac{2^{2}}{1^{2}}

- x = - R_{H} \times 4

x = 4 R_{H}

R_{H} = \frac{x}{4}

Step 2: Calculate energy for

{Be}^{3 +}

ion○ Expand

For ${Be}^{3 +}$ ion, the atomic number $Z = 4$ and the principal quantum number $n = 2$ . Using the same energy formula:

E_{2} = - R_{H} \frac{Z^{2}}{n^{2}}

E_{2} = - R_{H} \frac{4^{2}}{2^{2}}

E_{2} = - R_{H} \frac{16}{4}

E_{2} = - R_{H} \times 4

💡 Teacher's Secret Hint

Ensure correct atomic numbers and quantum numbers are used for each ion.

Step 3: Substitute

R_{H}

in terms of

x

○ Expand

Substitute the expression for $R_{H}$ from Step 1 ( $R_{H} = \frac{x}{4}$ ) into the energy equation for ${Be}^{3 +}$ :

E_{2} = - (\frac{x}{4}) \times 4

E_{2} = - x J

💡 Teacher's Secret Hint

Double-check the algebraic substitution to avoid errors.

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Chemistry Question 52 – NEET-UG 2024

Clue 1 of 3: Bohr's Model for Hydrogen-like Species

🧠 Full Solution Path