AP EAMCET · Chemistry · Structure of Atom
The energy of an electron in an orbit of hydrogen like ion with an orbit radius of \(52.9 \mathrm{pm}\) in \(\mathrm{J}\) is (ground state energy of electron in hydrogen atom is \(=-2.18 \times 10^{-18} \mathrm{~J}\) )
- A \(-4.36 \times 10^{-18}\)
- B \(-1.09 \times 10^{-17}\)
- C \(-8.72 \times 10^{-18}\)
- D \(-6.54 \times 10^{-18}\)
Answer & Solution
Correct Answer
(C) \(-8.72 \times 10^{-18}\)
Step-by-step Solution
Detailed explanation
Given, Orbit radius \(=52.9 \mathrm{pm}\) Ground state energy of electron in hydrogen atom…
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