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
b
d
b given
a
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
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
find force by derivating pe, equate to mv^2/r and then apply bohr quantisation, aka nL = 2pir.
L = h/p
already found while calc 4
-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
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