The orbital angular momentum of electron in the `n_(1)th` shell of element of atomic number `Z_(1) is L_(1)` an the same in the `n_(2)th` shell of ele

Correct Answer - B The transition from higher to first orbit causes appearance of Lyman line

According to (n+1) rule governing the filling of electrons in a multielectron atom, for 4s orbital, n+l=4+0=4 for 3d orbital,m+l=3+2=5. Therefore, 4s orbital fille dbefore electron filling takes place in...

Correct Answer - A::C a,c are the correct answers.

Since `E_(K)-E_(L)=(hc)/(lambda_(K_(alpha)))` It is given that `lambda_(K_(alpha))=21.3"pm" =0.0213 "nm"` Also, `" " E(eV) =(1242)/(lambda(nm))=(1242)/(0.0213)=58.3 eV` `rArr" " E_(K)-E_(L)=E` `rArr" " E_(K)-12.5=58.3` `rArr" "E_(K)=70.8 keV` `:.` Acceleration potential =70.8 keV

It is given that, `E_(K)=35.2keV,E_(L)=5.25keV,E_(M)=0.55 keV` `rArr" " E_(K)-E_(L)=(hc)/(lambda_(K_(A_(alpha))))=hv_(K_(alpha))` `rArr(35.2-5.25)xx10^(3)xx1.6xx10^(-19)=6.6xx10^(-34)v_(K_(alpha))` `rArr" " v_(K_(alpha))=(29.95xx10^(3)xx1.6xx10^(-19))/(6.6xx10^(-34))=7.2xx10^(-34)Hz` Similarly, `E_(K)-E_(M)=hv_(K_(beta))` `rArr" " v_(K_(beta))=((35.2-0.55)xx10^(3)xx1.63xx10^(-19))/(6.6xx10^(-34))=8.4xx10^(18)Hz` Also, `E_(L)-E_(M)=hv_(k_(alpha))` `rArr" " ((5.25-0.55)xx10^(3)xx1.6xx10^(-19))/(6.6xx10^(-34))=1.1xx10^(18)Hz`

Correct Answer - D `a=omega^(2)r` As `" " omegaprop(1)/(n^(3))" " and " : r propn^(2)` `" "aprop(1)/(n^(4))` `rArr" " (a_(M))/(a_(L))=((2)/(3))^(4)=(16)/(81)`

Correct Answer - B `L=n((h)/(2pi))i.e.,Lpropn.` In ground state n=1 so L= minimum.

Correct Answer - A The element which present just below the given element will have outermost electromic configuration as `4s^(2) 4p^(3),` so its full electronic configuration is `1s^(2),2s^(2),2p^(6),3s^(2)3p^(6),4s^(2), 3d^(10), 4p^(3)` and...

Correct Answer - D Cation at `3^(@) gt 2^(@) gt 1^(@)`

Orbital angular momentum `=sqrt(l(l+1))(h)/(2pi)` for p subshell, l=1 `=sqrt(2)(h)/(2pi)=(h)/sqrt(2pi)`