The number of s-electrons present in an ion with 55 protons in its unipositive state is
To count s-electrons, include all electrons in the s-subshells up to the element’s configuration.
The element with 55 protons is cesium (Cs), and its atomic number is Z = 55. The electronic configuration of neutral Cs is:
\(Cs : [Xe] 6s ^1\)
For the unipositive ion Cs+, it loses the single 6s electron, and the configuration becomes:
\(Cs^+: [Xe]\)
Now count the total number of s-electrons in\( Cs^+:\)
• \(1s^2, 2s^2, 3s^2, 4s^2, 5s^2\) contribute \(2 × 5 = 10\) s-electrons.
Thus, the number of s-electrons is 10.
Considering Bohr’s atomic model for hydrogen atom :
(A) the energy of H atom in ground state is same as energy of He+ ion in its first excited state.
(B) the energy of H atom in ground state is same as that for Li++ ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of He+ ion for its ground state.
(D) the energy of He+ ion in its first excited state is same as that for Li++ ion in its ground state.
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.