(a)
Electronic configuration is given as follows:
![[Kr]4d^{3}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B3%7D)
Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.
Thus, electronic configuration of neutral atom is
.
The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.
This corresponds to element: molybdenum. Thus, the tripositive atom will be
.
(b) The given electronic configuration is
.
The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.
This corresponds to element zirconium, represented by symbol Zr.
Answer:
4 x 10⁻⁴ J
Explanation:
C = 5000 pF, V = 400 V
Energy = CV²/2 = 5000 x 10⁻¹² x 400²/2 = 4 x 10⁻⁴ J
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