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Structure of Atom question

2017 · Shift 0 · Q23
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Structure of Atom question

2017 · Shift 0 · Q23

JEE MainChemistryStructure of AtomMCQ+4 / −1
The radius of the second Bohr orbit for hydrogen atom is: (Planck’s Const. h = 6.6262 × 10-34 Js; mass of electron = 9.1091 × 10-31 kg; charge of electron (e) = 1.60210 × 10-19 C; permittivity of vacuum ({varepsilon _0}) = 8.854185 × 10-12 kg-1 m-3 A2)
  1. A
    4.76 Ao\mathop {\rm A}\limits^oAo
  2. B
    2.12 Ao\mathop {\rm A}\limits^oAo
  3. C
    0.529 Ao\mathop {\rm A}\limits^oAo
  4. D
    1.65 Ao\mathop {\rm A}\limits^oAo
View written solutionFree

Correct answer: B

  1. Bohr radius formula

For hydrogen atom, the radius of the nnnth Bohr orbit is

rn=n2h2ε0πme2r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m e^2}rn​=πme2n2h2ε0​​

For the second orbit, n=2n=2n=2:

r2=4⋅h2ε0πme2r_2=4\cdot \frac{h^2 \varepsilon_0}{\pi m e^2}r2​=4⋅πme2h2ε0​​
  1. Substitute the given values

Given:

h=6.6262×10−34 J sh=6.6262\times 10^{-34}\ \text{J s}h=6.6262×10−34 J s m=9.1091×10−31 kgm=9.1091\times 10^{-31}\ \text{kg}m=9.1091×10−31 kg e=1.60210×10−19 Ce=1.60210\times 10^{-19}\ \text{C}e=1.60210×10−19 C ε0=8.854185×10−12\varepsilon_0=8.854185\times 10^{-12}ε0​=8.854185×10−12

So,

r2=4⋅(6.6262×10−34)2(8.854185×10−12)π(9.1091×10−31)(1.60210×10−19)2r_2=4\cdot \frac{(6.6262\times 10^{-34})^2(8.854185\times 10^{-12})}{\pi(9.1091\times 10^{-31})(1.60210\times 10^{-19})^2}r2​=4⋅π(9.1091×10−31)(1.60210×10−19)2(6.6262×10−34)2(8.854185×10−12)​
  1. Calculate step by step

First,

h2=(6.6262×10−34)2≈4.3907×10−67h^2=(6.6262\times 10^{-34})^2\approx 4.3907\times 10^{-67}h2=(6.6262×10−34)2≈4.3907×10−67

Then,

h2ε0≈(4.3907×10−67)(8.854185×10−12)≈3.888×10−78h^2\varepsilon_0\approx (4.3907\times 10^{-67})(8.854185\times 10^{-12}) \approx 3.888\times 10^{-78}h2ε0​≈(4.3907×10−67)(8.854185×10−12)≈3.888×10−78

Multiply by 444:

4h2ε0≈1.555×10−774h^2\varepsilon_0\approx 1.555\times 10^{-77}4h2ε0​≈1.555×10−77

Now,

e2=(1.60210×10−19)2≈2.5667×10−38e^2=(1.60210\times 10^{-19})^2\approx 2.5667\times 10^{-38}e2=(1.60210×10−19)2≈2.5667×10−38

Then,

me2≈(9.1091×10−31)(2.5667×10−38)≈2.338×10−68m e^2\approx (9.1091\times 10^{-31})(2.5667\times 10^{-38}) \approx 2.338\times 10^{-68}me2≈(9.1091×10−31)(2.5667×10−38)≈2.338×10−68

Now multiply by π\piπ:

πme2≈3.1416×2.338×10−68≈7.344×10−68\pi m e^2\approx 3.1416\times 2.338\times 10^{-68} \approx 7.344\times 10^{-68}πme2≈3.1416×2.338×10−68≈7.344×10−68

Therefore,

r2≈1.555×10−777.344×10−68≈2.12×10−10 mr_2\approx \frac{1.555\times 10^{-77}}{7.344\times 10^{-68}} \approx 2.12\times 10^{-10}\ \text{m}r2​≈7.344×10−681.555×10−77​≈2.12×10−10 m
  1. Convert into angstrom

Since

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

we get

r2≈2.12 A˚r_2\approx 2.12\ \text{\AA}r2​≈2.12 A˚
  1. Match with options
Option B: 2.12 A˚\boxed{\text{Option B: } 2.12\ \text{\AA}}Option B: 2.12 A˚​
  1. Check with stored correct answer

Stored correct answer is B, which matches our result.

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