Advertisements
Advertisements
प्रश्न
Calculate the energy required for the process.
\[\ce{He^+_{(g)} -> He^2+_{(g)} + e^-}\]
The ionisation energy for the H atom in its ground state is −13.6 ev atom−1.
Advertisements
उत्तर
\[\ce{He^+_{(g)} -> He^2+_{(g)} + e^-}\]
En = `(-13.6"z"^2)/"n"^2`
E1 = `(-13.6(2)^2)/(1)^2` = −54.4
E∞ = `(-13.6(2)^2)/(∞)^2` = 0
∴ Required Energy for the given process = E∞ – E1 = 0 – (– 54.4) = 54.4 ev
APPEARS IN
संबंधित प्रश्न
Electronic configuration of species M2+ is 1s2 2s2 2p6 3s2 3p6 3d6 and its atomic weight is 56. The number of neutrons in the nucleus of species M is
The electronic configuration of Eu (Atomic no. 63) Gd (Atomic no. 64) and Tb (Atomic no. 65) are:
The maximum number of electrons in a sub shell is given by the expression
If n = 6, the correct sequence for filling of electrons will be,
Consider the following electronic arrangements for the d5 configuration.
| ⥮ | ⥮ | 1 | ||
| (a) | ||||
| 1 | 1 | 1 | ⥮ | |
| (b) | ||||
| 1 | 1 | 1 | 1 | 1 |
| (c) | ||||
- which of these represents the ground state.
- which configuration has the maximum exchange energy.
what are the n and l values for 3px and `"4d"_("x"^2 - "y"^2)` electron?
Calculate the uncertainty in position of an electron, if Δv = 0.1% and υ = 2.2 × 106 ms-1
For each of the following, give the sub level designation, the allowable m values and the number of orbitals
- n = 4, l = 2
- n = 5, l = 3
- n = 7, l = 0
Give the electronic configuration of Mn2+ and Cr3+
An atom of an element contains 35 electrons and 45 neutrons. Deduce
- the number of protons
- the electronic configuration for the element
- All the four quantum numbers for the last electron
