Home › JEE Advanced › Chemistry › Atomic combinations › For Bohr's hydrogen-like atom, what is the veloc…
For Bohr's hydrogen-like atom, what is the velocity of electron in n-th orbit (in terms of fine structure constant α and c)?
Av_n = c / Z
Bv_n = c α Z / n (where α ≈ 1/137)
Cv_n = c
Dv_n = c × n
Answer & Solution
Correct answer: B. v_n = c α Z / n (where α ≈ 1/137)
v_n = (1/n) × (Zα c) where α = e²/(4πε₀ ħc) ≈ 1/137 is the fine structure constant. For H ground state: v₁ ≈ c/137 ≈ 2.19 × 10⁶ m/s. Relativistic effects small for light atoms, but become significant for heavy atoms (high Z).
Related questions
The strength of a bond depends chiefly on the extent of:Sidewise overlapping produces charged clouds shaped like a:In a pi bond, the axes of the overlapping orbitals remain:Head-on overlap along the internuclear axis is also called:A sigma bond is formed by overlap that is:In a molecule such as HF, the covalent bond is:A bond where the electron pair sits exactly between identical nuclei is:The nitrogen bond enthalpy of 946 kJ per mol is one of the: