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NVAEasyJEE 2023Bohr's Model & Hydrogen Spectrum

JEE Physics 2023 Question with Solution

The radius of the fifth orbit of Li++^{++} is _____ ×10−12 m\times 10^{-12} \, \text{m}. Take the radius of hydrogen atom = 0.51 A˚0.51 \, \text{\AA}.

Answer

Correct answer:425

Step-by-step solution

Standard Method

Given: Radius of hydrogen atom = 0.51 A˚0.51 \, \text{\AA}, for Li++^{++} we have Z=3Z = 3, and the required orbit is the fifth orbit so n=5n = 5.

Find: The value in _____ ×10−12 m\times 10^{-12} \, \text{m} for the radius of the fifth orbit.

Using Bohr's model, the radius of the nnth orbit of a hydrogen-like atom is

rn=0.51n2Z A˚r_n = \frac{0.51n^2}{Z} \, \text{\AA}

For Li++^{++}, substitute Z=3Z = 3 and n=5n = 5:

r5=0.51×523 A˚=0.51×253 A˚r_5 = \frac{0.51 \times 5^2}{3} \, \text{\AA} = \frac{0.51 \times 25}{3} \, \text{\AA}

Now convert angstrom to metre:

r5=0.51×253×10−10 mr_5 = 0.51 \times \frac{25}{3} \times 10^{-10} \, \text{m} =17×25×10−12 m= 17 \times 25 \times 10^{-12} \, \text{m} =425×10−12 m= 425 \times 10^{-12} \, \text{m}

Therefore, the required numerical value is 425.

Bohr Model Expansion

Given: A hydrogen-like ion Li++^{++} with atomic number Z=3Z = 3, hydrogen atom radius 0.51 A˚0.51 \, \text{\AA}, and orbit number n=5n = 5.

Find: Radius of the fifth orbit in the form _____ ×10−12 m\times 10^{-12} \, \text{m}.

According to Bohr's model, the orbital radius is

rn=0.51n2Z A˚r_n = \frac{0.51n^2}{Z} \, \text{\AA}

where nn is the principal quantum number and ZZ is the atomic number.

Substituting for Li++^{++}:

r5=0.51×253 A˚r_5 = \frac{0.51 \times 25}{3} \, \text{\AA}

Writing 1 A˚=10−10 m1 \, \text{\AA} = 10^{-10} \, \text{m},

r5=0.51×253×10−10 mr_5 = 0.51 \times \frac{25}{3} \times 10^{-10} \, \text{m}

Since 0.51/3=0.170.51/3 = 0.17,

r5=0.17×25×10−10 mr_5 = 0.17 \times 25 \times 10^{-10} \, \text{m} =4.25×10−10 m= 4.25 \times 10^{-10} \, \text{m} =425×10−12 m= 425 \times 10^{-12} \, \text{m}

Therefore, the radius of the fifth orbit of Li++^{++} is 425×10−12 m425 \times 10^{-12} \, \text{m}, so the answer is 425.

Common mistakes

  • Using the hydrogen formula without dividing by ZZ is incorrect because Li++^{++} is a hydrogen-like ion with nuclear charge Z=3Z = 3. Always use rn∝n2Zr_n \propto \frac{n^2}{Z} for hydrogen-like species.

  • Taking Li++^{++} as neutral lithium is wrong because the ion has only one electron and follows the hydrogen-like Bohr model. Use atomic number Z=3Z = 3, not electron count of neutral lithium.

  • Missing the unit conversion from angstrom to metre leads to the wrong power of 1010. Remember that 1 A˚=10−10 m1 \, \text{\AA} = 10^{-10} \, \text{m} before matching the asked form ×10−12 m\times 10^{-12} \, \text{m}.

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