The ionization energy of H-atom is `13.6eV`. Calculate the is ionization energy of `Li^(+2)`ion-
A. `6hcR`
B. `2hcR`
C. `9hcR`
D. `hcR`
1 Answers
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Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in gruond state i
Correct Answer - A Excitation energy , `DeltaE =E_(2)-E_(1)=13.6 [(1)/(1^(2))-(1)/(2^(2))]` `" " 40.8=13.6 Z^(2)xx(3)/(4)` `:. " " Z=2` So, required energy to remove the electron from gound state `=+(13.6Z^(2))/((1)^(2))=13.6(2)^(2)=54.4 eV`
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Match the following : B.E.-Binding energy I.E.-Ionization energy `{:("Column-I","Column-II"),((A)"B.E. of"He^(+)"atom in an excited state" ,(p)"Infrar
Correct Answer - A-q,r ; B-p,s ; C-p,s,t ; D-q (A)B.E. of `He^+` atom =`(13.6xx2^2)/n^2` n=1,2,3… Hence it can be 13.6 ev , 3.4 ev both ( B)In `7to3` transition `Deltan=7-3=4`...
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The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called as first ionization energy `(IE_1)`.Si
Correct Answer - D Orbitals bearing lower value of n will be more closer to the nucleus and thus electrons will experience greater attraction from nucleus and so its removal will...
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Statement-1:Cuprous ion `(Cu^+)` is colourless whereas cupric ion `(Cu^(2+))` produce blue colour in solution. Statement-2:Cupros ion `Cu^+` has unpai
Correct Answer - C Cupric ion proudces blue colour in solution due to the presence of unpaired electrons (d-d transition) while cuprous ion does not have unpaired electron.
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The ionization potential for hydrogen atom is `13.6 eV`, the ionization potential for `He^(+)` is
Correct Answer - A I.P of `He^(+)` = I.P. of `H xx Z^(2)` (Where Z is the atomic number of He) = 13.6 eV `xx (2)^(2)` = 54.4 eV.
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If `13.6eV` energy is required to separate a hydrogen atom into a proton and an electron, then the orbital radius of electron in a hydrogen atom is
Correct Answer - A Here, `E=-13.6eV=-13.6xx1.6xx10^(-19)=-2.2xx10^(-18)J` `E=(-e^2)/(8piepsilon_0r)` `therefore` As orbital radius, `r=(-e^2)/(8piepsilon_0E)=(9xx10^(9)xx(1.6xx10^(-19))^2)/(2xx(2.2xx10^(-18)))=5.3xx10^(-11)m`.
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The ground state energy of hydrogen atom is -13.6eV. What are P.E. and K.E. of electron in this state?
Correct Answer - C Here the ground state energy , (E) `=-13.6eV` since, K.E. of the electron `=-13.6eV` `=13-6xx1.6xx10^(-19)J=21.76xx10^(-19)J=2.18xx10^(-18)J`.
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The ionisation energy of `H` atom is `21.79xx10^(-19) J`. The the value of binding energy of second excited state of `Li^(2+)` ion
Correct Answer - B B.E.`=(21.79xx10^(-19)(3)^(2))/((3)^(2))=21.79xx10^(-19)J`
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Energy of H-atom in the ground state is `-13.6eV`. Hence energy in the second excited state is
Correct Answer - 3 `2^(nd)` excited state will be the `3^(rd)` energy level. `En=13.6/n^(2) eV` or `E=13.6/9=1.51 eV`.
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A hydrogen atom in ground state is moving with a kinetic energy of 30 eV. It collides with a deuterium atom in ground state at rest. The hydrogen atom
Correct Answer - `20 eV, 13.2 eV`
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