Atomic

c

For 1s, l = 0 and thus if you remember the quantum eqn for psi, there were three terms dependent on radial distance

sigma^l, a polynomial term of the order n-l-1 and e^-sigma/2

in 1s, l = 0, n-l-1=0

thus only e^-sigma/2 term is left

As sigma is proprtional to r, psi decreases as r increases

thus, probability is max at r=0 and decreases exponentially to 0 at infinity

a,b

spherical nodes are radial nodes

b,d

1. c

2. c

b

d

b given

b

a

b

b

light is emitted/absorbed at every step as in atoms, all orbitals are no degenrate and they follow n+l rule

so just find the transitions where light is EMITTED

c ha ans

ans at the back of neeraj is wrong

d

b

d

as l =3, n is greater than equal to 4

thus make cases and see what fits and what doesnt

induced means the wavelengths released on de-excitation

a,b,c

since all photons absorbed have the same energy, we know that there is only on transition.

since n photos are released, we know each atom only underwent 1 transition as otherwise more than n photos wouldve been released

hence transition is 1->2

Thus ans is multiple of 10.2 ev

a,b,d

4 c

5 d

6 c

1. find force by derivating pe, equate to mv^2/r and then apply bohr quantisation, aka nL = 2pir.

   L = h/p

2. already found while calc 4

3. -2*KE in nth orbit

14 b

15 c

16 d

Electrons after taking 2.7 ev go to n = 4 (because 6 photos emitted on deexcitation)

As energy of photos is both more, less and equal to 2.7 ev, they couldnt have come from n= 3

since it is an excited state after all, initial n = 2

solve the rest

more a question of modern rather than chem but ok

17 b

18 d

19 c

just for practice

35 a

36 b

37 c