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Physics Question 43 – JEE-MAIN 2025

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Considering Bohr's atomic model for hydrogen atom : (A) the energy of H atom in ground state is same as energy of He+ ion in its first excited state. (B) the energy of H atom in ground state is same as that for Li++ ion in its second excited state. (C) the energy of H atom in its ground state is same as that of He+ ion for its ground state. (D) the energy of He+ ion in its first excited state is same as that for Li++ ion in its ground state. Choose the correct answer from the options given below :

Recall the formula for the energy of an electron in the n-th orbit of a hydrogen-like atom.

🥷
Ninja StrategyDirect Evaluation and Elimination

Evaluate each statement (A, B, C, D) using the Bohr energy formula. Once false statements are identified, eliminate options that contain them.

Video Walkthrough
Step 1: Recall Bohr's Energy Formula✦ Active

The energy of an electron in the n-th orbit of a hydrogen-like atom with atomic number Z is given by:

En=13.6Z2n2 eV
Step 2: Evaluate Each Statement○ Expand

Let's calculate the energy for each specified state:

(A) H atom (ground state): Z=1,n=1EH(n=1)=13.61212=13.6 eV
He+ ion (first excited state): Z=2,n=2EHe+(n=2)=13.62222=13.6 eV

Statement (A) is TRUE as EH(n=1)=EHe+(n=2).

(B) H atom (ground state): EH(n=1)=13.6 eV
Li++ ion (second excited state): Z=3,n=3ELi++(n=3)=13.63232=13.6 eV

Statement (B) is TRUE as EH(n=1)=ELi++(n=3).

(C) H atom (ground state): EH(n=1)=13.6 eV
He+ ion (ground state): Z=2,n=1EHe+(n=1)=13.62212=54.4 eV

Statement (C) is FALSE as EH(n=1)EHe+(n=1).

(D) He+ ion (first excited state): EHe+(n=2)=13.6 eV
Li++ ion (ground state): Z=3,n=1ELi++(n=1)=13.63212=122.4 eV

Statement (D) is FALSE as EHe+(n=2)ELi++(n=1).

💡 Teacher's Secret Hint

Pay close attention to the atomic number Z and the principal quantum number n for each species and state.

Step 3: Identify Correct Option○ Expand

Only statements (A) and (B) are correct. Therefore, the option that includes (A) and (B) only is the correct answer.

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