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

c

L = h/p

dL/L = dp/p

Thus dL = L* dp/p

Find dp/p using uncertainty principle

a,b

orbital L = h/2pi * root (l(l+1))

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 not 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

c

Faraday mi goat confirmed first

c

this is because in anode rays, more often than not, cations are involved as all electrons arent ionised

thus its positive residue of atoms, not nuclei

c

d

c

Lenard

broski didnt give a number

he just said ‘almost all’

hence a is wong

a

We equate Ke and Pe and thus q/m remains same for both

b

inversely proportional to sin^4 (theta/2)

b

its just the wavelentgh

a

trapping the signal basically measn that minimum this amount of energy must fall on the eyes to make sure a signal is sent to the brain

Here, unlike phy theyre only taking actually emitted photons, not reflected ones

actually theyre assuming a=1

ans is hence c

c

Arvin ash ftw

ans is c

just do the maths habibi aur kuch nahi kar sakte

c

hopefully this was easy and you didnt do silly mistake

it becomes halved

derive everything it doesnt really take all that long

2nd IE= 54.4 ev

and this needs to be greater than 13.6 obv cuz wo hydrogen mei hoti hai and not only is this electron closer, nucleus has more charge too

so ans is between 13.6 and 54.4

We cant find value khud se

hence d

c

Rydberg’s constant is proportional to mass of electron

a

q1 gets cancelled fam

R is proportional to e^4

hence ans is d

It feels like it should be e^2, but keep in mind that radius is inversely proprotional to e^2 too

so R is proportional to q^2/r which is proportional to q^4

a

there’s only 1 atom and there’s only 1 transition

hence 1 photon only

d

eay ques hai

d

10 to 4 gives 6 lines in brackett

b

nodal planes are essentially angular nodes only

so number of nodal planes = l

and is b GOTCHA

Number of nodal planes = angular nodes

number of nodal surfaces = total number of nodes = n-1

l = 3

thus there are 2(3)+1 = 7 orbitals

Spin multiplicity is 2s+1

Thus max is 8 when we have d7

min in 1 when its fully filled so its 2(0)+1

ans is a

ANY fully filled or half filled orbital has spherical symmetry

ANY means s,p,d,f,g and so on

if they’re FULLY OR HALF filled then spherical symmetry