The quantum number of six electrons are given below ,Arrange them in order of increasing energies . If any of these combination has have the same energy lists : (1) n-4 ,l=2 , m_(1) = 2m_(1) = (1)/(2) (2) n=3 , l = 2, m_(1) = 1 ,m_(s) = (1)/(2) (3 ) n=4 , l = 1 , m_(1) = =0 m_(s ) = (1)/(2) (4 ) n=3: l = 2 m_(1) = - 2 m_(s ) = - (1)/(2) (5 ) n=3 , l = 1,m_(1) = - 1 m_(s ) = (1)/(2) (6) n=4 , l = 1, m_(1) = 0 m_(s ) = (1)/(2)

Compute (3m -2n) ^(3)

For as Ap. S_(n)= m and S_(m) = n . Prove that S_(m+n)=-(m+n).(m ne n)

Explain giving reasons which of the following sets of quantum number are not possible (a ) n=0, l =0 m_(l) = 0, m_(s ) =+ (1)/(2) ( b) n=1 , l = 0 m_(l) = 0, m_(s ) = - (1)/(2) ( c) n=1 , l = 1, m_(l ) = 0, m_(s ) = + (1)/(2) (d ) n= 2, l = 1, m_(l ) = 0, m_(s ) = (1) /(2) ( e) n=3, l = 3, m_(l) = 3, m_(s ) = + (1)/(2) (f ) n=3, l = 1, m_(l) = 0, m_(s) l = + (1)/(2)

Arrange in increasing order of energy (I ) n=2 ,l =0 , m_(1) = 0 m_(s ) = (1)/(2) (ii) n= 2 , l = 1 , m_(1) =+ 1 , m_(s ) = (1)/(2) (ii) n=3 ,l m_(1) =+ 1 ,m_(s) = (1)/(2) (iv) n=3,l = m_(l) =-1 m_(s ) =+ (1)/(2)

Two moving coil meters M_1 and M_2 have the following particulars : R_1 = 10 Omega , N_1 = 30 A_1 = 3.6 xx 10^(-3) m^2 , B_1 = 0.25 T R_2 = 14Omega , N_2 = 42 A_2 = 1.8 xx 10^(-3) m^2 , B_2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_2 and M_1 .

If L= {1, 2, 3, 4}, M= {3, 4, 5, 6} and N= {1, 3, 5}, then verify that L - (M cup N)= (L - M) cap (L - N) .

A biard is at a point P(4m,-1m,5m) and sees two points P_(1) (-1m,-1m,0m) and P_(2)(3m, -1m,-3m) . At time t=0 , it starts flying in a plane of the three positions, with a constant speed of 2m//s in a direction perpendicular to the straight line P_(1)P_(2) till it sees P_(1) & P_(2) collinear at time t. Find the time t.

If ((2n),(3)) = 11 ((n),(3)) then n = .....

If three lines whose equations are y=m_(1) x+c_(1), y=m_(2) x + c_(2) and y=m_(3) x + c_(3) are concurrent, then show that m_(1) (c_(2) - c_(3))+m_(2) (c_(3) - c_(1) ) + m_(3) ( c_(1) - c_(2) ) =0 .