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+4 marks−1 if incorrectNumericalPrevious-year question

Bohr Orbit Radius of Hydrogen-like Ions

The radius of fifth orbit of the Li^(++) is ______ × 10^(-12)m. Take: radius of hydrogen atom = 0.51AA

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Correct answer

The radius of the fifth orbit of ${L i}^{+ +}$ is $425 \times 10^{- 12} m$ .

Option analysis

Why each option works or fails

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Solution

StepWorking

01Given

Radius of ground state hydrogen atom $r_{0} = 0.51 \overset{˚}{A} = 0.51 \times 10^{- 10} m$ , principal quantum number $n = 5$ , atomic number for ${L i}^{+ +}$ is $Z = 3$ .

02Find

The radius of the 5th orbit of ${L i}^{+ +}$ expressed in the form $N \times 10^{- 12} m$ , finding the integer $N$ .

03Strategise

Use Bohr's formula for the orbital radius of a hydrogen-like atom: $r_{n} = r_{0} \frac{n^{2}}{Z}$ . Substitute $n = 5$ and $Z = 3$ , then convert units from Angstroms to meters with power $10^{- 12} m$ .

04Execute

$r_{5} = 0.51 \times \frac{5^{2}}{3} = 0.51 \times \frac{25}{3} = 0.17 \times 25 = 4.25 \overset{˚}{A} = 425 \times 10^{- 12} m$ . Thus, the required value is 425.

✓Verify

Check dimensions and scaling: $r \propto n^{2} / Z$ . For $n = 5$ , factor is $25 / 3 \approx 8.33$ . $0.51 \times 8.33 \approx 4.25 \overset{˚}{A} = 425 pm = 425 \times 10^{- 12} m$ . The scaling and unit exponent match.

Hints that build this answer step by step

What formula gives the radius $r_{n}$ of the $n$ -th Bohr orbit of a hydrogen-like ion in terms of the ground state radius of hydrogen $r_{0}$ ?
$r_{n} = r_{0} \frac{n^{2}}{Z}$
For the fifth orbit of ${L i}^{+ +}$ , what values of $n$ and $Z$ should be substituted into $r_{n} = r_{0} \frac{n^{2}}{Z}$ with $r_{0} = 0.51 \overset{˚}{A}$ ?
$n = 5$ , $Z = 3$ , giving $r_{5} = 0.51 \times \frac{25}{3} \overset{˚}{A} = 4.25 \overset{˚}{A}$
Convert $4.25 \overset{˚}{A}$ into the requested format: $____\times 10^{- 12} m$ .
$425$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why do we take Z = 3 for Li++?

Z is the atomic number (number of protons in the nucleus), which is 3 for Lithium. The ++ indicates two electrons are removed, leaving a 1-electron system to which Bohr model applies.

Answer

Accepted answer: 425

The radius of the fifth orbit of ${L i}^{+ +}$ is $425 \times 10^{- 12} m$ .

Why each option works or fails

Step-by-step solution

Given: Radius of ground state hydrogen atom $r_{0} = 0.51 \overset{˚}{A} = 0.51 \times 10^{- 10} m$ , principal quantum number $n = 5$ , atomic number for ${L i}^{+ +}$ is $Z = 3$ .
Find: The radius of the 5th orbit of ${L i}^{+ +}$ expressed in the form $N \times 10^{- 12} m$ , finding the integer $N$ .
Strategise: Use Bohr's formula for the orbital radius of a hydrogen-like atom: $r_{n} = r_{0} \frac{n^{2}}{Z}$ . Substitute $n = 5$ and $Z = 3$ , then convert units from Angstroms to meters with power $10^{- 12} m$ .
Execute: $r_{5} = 0.51 \times \frac{5^{2}}{3} = 0.51 \times \frac{25}{3} = 0.17 \times 25 = 4.25 \overset{˚}{A} = 425 \times 10^{- 12} m$ . Thus, the required value is 425.
Verify: Check dimensions and scaling: $r \propto n^{2} / Z$ . For $n = 5$ , factor is $25 / 3 \approx 8.33$ . $0.51 \times 8.33 \approx 4.25 \overset{˚}{A} = 425 pm = 425 \times 10^{- 12} m$ . The scaling and unit exponent match.

Bohr Orbit Radius of Hydrogen-like Ions

What feels right?

The radius of the fifth orbit of ${L i}^{+ +}$ is $425 \times 10^{- 12} m$ .

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

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