WBJEE · Chemistry · Structure of Atom
The radius of the first Bohr orbit of a hydrogen atom is \(0.53 \times 10^{-8} \mathrm{~cm}\). The velocity of the electron in the first Bohr orbit is
- A \(2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}\)
- B \(4.376 \times 10^{8} \mathrm{~cm} \mathrm{~s}^{-1}\)
- C \(1.094 \times 10^{8} \mathrm{~cm} \mathrm{~s}^{-1}\)
- D \(2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(A) \(2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
Hint: \(r=0.53 \times 10^{-9} \mathrm{~cm}=0.53 \times 10^{-9} n^{2}\) \(\therefore \mathrm{n}=1\) \(\mathrm{~V}=2.18 \times 10^{9} \mathrm{cm} / \mathrm{s}\)
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