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JEE Advanced · Chemistry · 2. Atomic Structure

According to Bohr's theory \(E_n=\) Total energy, \(K_n=\) Kinetic energy, \(V_n=\) Potential energy, \(r_n=\) Radius of \(n\)th orbit

  1. A (A) R, (B) Q, (C) P, (D) Q
  2. B (A) S, (B) R, (C) R, (D) S
  3. C (A) R, (B) Q, (C) P, (D) S
  4. D (A) S, (B) P, (C) R, (D) Q
Verified Solution

Answer & Solution

Correct Answer

(C) (A) R, (B) Q, (C) P, (D) S

Step-by-step Solution

Detailed explanation

According to Bohr's theory,
Total energy is \(E_n=K_n+V_n\)
Kinetic energy \(=K_n=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}\)
Potential energy \(=V_n=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}\)
\(\therefore \quad E_n=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}\)
Radius of \(n\)th orbit \(\left(r_n\right)=\frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}\)
\(E_n=-\frac{m e^4}{8 \varepsilon_0^2 h^2} \times \frac{1}{n^2} \)
(A) \(\frac{V_n}{K_n}=\frac{-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}}{\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}}=-2\)
Hence, \((A)\) match with \((R)\).
(B) \(E_n \propto \frac{1}{n^2}\) or \(E_n \propto \frac{1}{r_n}\)
Radius of \(n\)th orbit \(r_n \propto E_n^x\) \(\therefore \quad x=-1\)
Hence, (B) match with (Q).
(C) Angular momentum \(=\frac{h}{2 \pi} \sqrt{l(l+1) l}=0,1,2, \ldots\)
For the lower orbit \(n=1\)
\(\therefore \quad l=0\) and \(m=0\)
Hence, angular momentum of lowest orbit \(=\frac{h}{2 \pi} \sqrt{0(0 H)}=0\)
(C) match with (P)
(D) \(\frac{1}{r^n} \propto Z^y\) as \(r_n \propto \frac{1}{Z} \therefore y=1\)
Hence, (D) match with (S).
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