According to Bohr's model, the highest kinetic energy is associated with the electron in the:
Detailed Step-by-Step Solution
To determine the highest kinetic energy, we must use the energy relationships defined by Bohr's model for hydrogen-like species.
Step 1: The Kinetic Energy Formula
According to Bohr's model, the Total Energy (\(E_n\)) of an electron in the \(n^{\text{th}}\) orbit is given by:
We know that Kinetic Energy (\(KE\)) is equal to the negative of Total Energy (\(KE = -E_n\)). Therefore:
Thus, kinetic energy is directly proportional to \( \frac{Z^2}{n^2} \). We just need to calculate this ratio for each given option.
Step 2: Evaluate Each Option
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(A) First orbit of H atom:
Here \( Z = 1 \) and \( n = 1 \).
\( \text{Ratio } \left(\frac{Z^2}{n^2}\right) = \frac{1^2}{1^2} = \mathbf{1} \) -
(B) First orbit of He\(^+\):
Here \( Z = 2 \) and \( n = 1 \).
\( \text{Ratio } \left(\frac{Z^2}{n^2}\right) = \frac{2^2}{1^2} = \mathbf{4} \) -
(C) Second orbit of He\(^+\):
Here \( Z = 2 \) and \( n = 2 \).
\( \text{Ratio } \left(\frac{Z^2}{n^2}\right) = \frac{2^2}{2^2} = \frac{4}{4} = \mathbf{1} \) -
(D) Second orbit of Li\(^{2+}\):
Here \( Z = 3 \) and \( n = 2 \).
\( \text{Ratio } \left(\frac{Z^2}{n^2}\right) = \frac{3^2}{2^2} = \frac{9}{4} = \mathbf{2.25} \)
Conclusion: Comparing the values (1, 4, 1, and 2.25), the highest ratio is 4. Therefore, the highest kinetic energy is associated with the first orbit of He\(^+\). The correct option is (B).
Mastering Bohr Model Proportionalities
In JEE Advanced, many questions can be solved without calculating the exact numerical value by simply comparing proportionalities. Remembering how Energy, Velocity, and Radius scale with \(Z\) (Atomic Number) and \(n\) (Orbit Number) is a massive time-saver.
For example: \( v \propto \frac{Z}{n} \), \( r \propto \frac{n^2}{Z} \), and \( E \propto \frac{Z^2}{n^2} \).
To build absolute confidence in these relationships and tackle complex multiple-choice questions, we highly recommend reading our detailed conceptual guide on the Structure of Atom Class 11 Chemistry.
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