બિસ્મથનું ગ્રાઉન્ડ સ્ટેટ ઈલેક્ટ્રોનિક કન્ફિગરેશન (`Z=83`) છે

The ground state electronic configuration of bismuth ( Z=83 ) is

Atomic no. of X,Y &Z are 33,53 &83 respectively.

Solve : (14y)/(3) + (3)/(2) le (20y)/(3) - (83)/(4) , where y in Z

Element with Z = 83 belongs to which block?

The number of elements in the set {(a,b):2a^(2)+3b^(2)=83,a,binZ} , where Z is the set of all integers, is

find the number of elements which arae iosfiaphers of ""_(89_Ac^(227) An element is respresented as (A.Z) (231,91),(223,87),(227,90),(223,88_),(219,85),(215,83),(215,84),(207,82),(211,83),

The distance of the point (1,3,-7) from the plane passing through the point (1,-1,-1) , having normal perpendicular to both the lines (x-1)/(1)=(y+2)/(-2)=(z-4)/(3) and (x-2)/(2)=(y+1)/(-1)=(z+7)/(-1) is (5)/(sqrt(83)) (2) (10)/(sqrt(74)) (3) (20)/(sqrt(74)) (4) (10)/(sqrt(83))

The mass of nucleus ._(z)X^(A) is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of mass m_(1) and m_(2) only if (m_(1)+m_(2)) lt M . Also two light nuclei of massws m_(3) and m_(4) can undergo complete fusion and form a heavy nucleus of mass M ''only if (m_(3)+m_(4)) gt M ''. The masses of some neutral atoms are given in the table below. |{:(._(1)^(1)H,1.007825u,._(1)^(2)H,2.014102u,),(._(1)^(3)H,3.016050u,._(2)^(4)H,4.002603u,),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,),(._(30)^(70)Zn,69.925325u,._(34)^(82)Se,81.916709u,),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.97445u,),(._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u,):}| The correct statement is