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>> Solved Questions of Atomic Structure Part I
Why Bohr’s orbits are called stationary states?
This is because the energies of orbits in which the electrons revolve are fixed.
Explain why the electronic configuration of Cu is 3d104s1 and not 3d94s2.
In the 3d104s1 the d-sub shell is completely filled which is more stable.
Fe3+ ion is more stable than Fe2+ ion. Why?
In Fe3+ ion 3d sub shell is half filled hence more stable configuration.
Calculate the accelerating potential that must be applied to a proton beam to give it an effective wavelength of 0.005 nm.
v = h/mλ
ev = 1/2mv2
Putting the values we get
V = 32.85 volt
Give one example of isodiapheres.
Isodiapheres have same difference between the number of neutrons and protons. For example
3919K & 3115P
Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n = 6 to n = 4 transition in He+. [Neglect reduced mass effect].
v-He+ = RZ2 [ 1/42 - 1/62]
= 4R [ 36 - 16/36 x 16 ] = 5R/36
v-H = R x 12 [ 1/22 - 1/n2]
∴ v-He+ = v-H
5R/36 = R/4 - R/n2
On solving above equation
n2 = 9
∴n = 3
Or corresponding transition from 3 → 2 in Balmer series of hydrogen atom has same frequency as that of 6 → 4 transition in He+.
Calculate ionization potential in volts of (a) He+ and (b) Li2+
I.E. = 13.6Z2/n2
= 13.6 x Z2 \12 [Z =2 for He+]
= 13.6 x 4 = 54.4 eV
Similarly for Li2+ = 13.6 x 32/12
= 13.6 x 9 = 122.4 eV
Calculate the ratio of K.E and P.E of an electron in an orbit?
K.E. = Ze2/2r
P.E. = -Ze2/r
∴ P.E. = –2K.E
∴ K.E/P.E = - 1/2
How many spectral lines are emitted by atomic hydrogen excited to
nth energy level?
Thus the number of lines emitted from nth energy level
= 1 + 2 + 3 +………… n – 1 = ∑(n – 1)
∑n = n(n+1)/2
∴ ∑ (n – 1) = ( n-1) (n-1+1)/2 = (n-1) (n)/2
Number of spectral lines that appear in hydrogen spectrum when an electron jumps from nth energy level = n (n-1)/2