The energy of electron in the nth Bohr orbit of H-atom is …………
(A) \(-\frac{2.18\times 10^{-16}}{n^2}J\)
(B) \(-\frac{2.18\times 10^{-18}}{n}J\)
(C) \(-\frac{2.18\times 10^{-18}}{n^2}J\)
(D) \(-\frac{2.18\times 10^{-16}}{n}J\)
Answered Feb 05, 2023
Option : (C) \(-\frac{2.18\times 10^{-18}}{n^2}J\)
Correct Answer - `3.25xx10^(18) rev.//sec`
Correct Answer - `-2.304xx10^(-18) J`
Correct Answer - `+n xx L.C. `
Correct option is B) 2
Correct option is B) emitted
B) ionization energy
Kuch bhi
We Know that, r = 0.529 x \(\frac{n^2}{z}\)pm Where, n = Orbit Number (Principle Quantum Number) z = Atomic Number ∴ r = 0.529 x \(\frac{4^2}{1}\) = 0.529 x 4 x 4 = 8.464 pm
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