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Question-1

Calculate the frequency and energy associated with photon of radiations having a wavelength of 6200 Ao. Plank's constant = 6.625 ×10-27 ergs sec.

Solution:
We know that c = λ or =

The value of c for all electromagnetic radiation = 3.0 ×108 m sec-1.

λ = 6200 Ao = 6200 ×10-8 cm

  = 6200 ×10-10 m

  = 62 ×10-8 m

= = 4.839 ×1014 cycles sec-1.

The energy (E) associated with a radiation is given by

E = hυ

E = 6.625 ×10-27 ×4.839 1014

  = 2.914 ×10-12 ergs

  = Joules

  = 2.914 ×10-19 Joules.

Question-2

Calculate the wave number of lines having the frequency of 5 ×1016 cycles per sec.

Solution:
Given c = 3 ×108 m/sec

υ= 5 ×1016 cycles/sec

υ= ?

We know that

υ= = = 1.666 ×108 m-1.

Question-3

In a hydrogen atom, an electron jumps from a third orbit to the first orbit. Find out the frequency and wavelength of the spectral line.

Solution:
(i) When an electron jumps from a higher orbit n2 to the lower orbit n1, the frequency υof the radiation is given by

υ= 3.29 ×105 cycles sec-1

Here, n1 = 1 and n2 = 3

υ = 3.29 × 105 cycles sec-1

  = 3.29 ×105 cycles sec-1

  = 3.29 ×105 ×0.889 cycles sec-1

  = 2.925 ×1015 cycles sec-1

Now wavelength λ =

λ = = 1.0256 ×10-7 m

  = 1.0256 ×10-7 ×1010 Ao = 1025.6 Ao.

Thus the wavelength of light emitted falls in the UV region of the electromagnetic spectrum.

Question-4

RH = 1.09678 ×107 m-1, c = 3 ×108 ms-1, h = 6.625 ×10-34 Js.

 


Solution:
= RZ2

For lowest frequency in Lyman series

n1 = 1, n2 = 2

For H, Z = 1

= 1. 09678 ×107 ×1 =

λ = = 1215 ×10-10 m or 1215 Ao.

Again, c = λ

- = = = 0.002469 ×1018 Hz = 2.469 × 1015 Hz

Energy E = hυ

E = 6.625 ×10-34 ×2.469 ×1015 = 10.22 eV

For Li2+, Z = 3

Li2+= (3)2 ×10.22 = 9 ×10.22 = 91.98 eV.

Question-5

Calculate the uncertainty in the position of a particle when the uncertainty in the momentum is (a) 1 ×10-2 (b) zero.

Solution:
(a) According to the uncertainty principle,

Δx. Δp

Putting the values of

h = 6.62 ×10-34 Joules-sec

Δp = 1 ×10-7 Kg-m-sec-1

Δx ×10-7 =

Δx = m

    = 0.527 ×10-27 m

(b) We know that Δx =

When ΔP = 0, the denominator in the above expression becomes zero; hence the uncertainty in position becomes infinity.

Question-6

(iii) 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1.

 


Solution:
(a) electronic configuration of elements with atomic number

19 1s2, 2s2, 2p6, 3s2, 3p6, 4s1

28 1s2, 2s2, 2p6, 3s2, 3p6, 3d8, 4s2

29 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1

(b) (i) Atomic number of the element is 2+2+6+2+6+1=19

Therefore, the element is potassium.

(ii) Atomic number of the element is 2+2+6+2+6+5+1=24

Therefore, the element is chromium.

(iii) Atomic number of the element is 2+2+6+2+6+10+1=29

Therefore, the element is copper.

Question-7

An electron is in a 4f orbital. What possible values for the quantum numbers n, l, m and s can it have?

Solution:
For an electron in a 4f orbital,

n = 4, l = 3, m = -3, -2, -1, 0, +1, +2, +3, s = + and - for each value of m.

Question-8

A neutral atom has 2K, 8L, 5M electrons. Find out the following from the data:

(a) atomic number,

(b) total number of s electrons,

(c) total number of p electrons,

(d) number of protons in the nucleus, and

(e) valency of element.

Solution:
(a) Atomic number = No. of protons = No. of electrons

Total no. of electrons = 2 + 8 + 5 = 15

Hence atomic number = 15

(b) Total number of s electrons. To find out it, we are to write electronic configuration of At. No. = 15

1s2, 2s2, 2p6, 3s2, 3p3

Total electrons = 6

(c) Total number of p electrons = 9

(d) Number of protons in the nucleus = Number of electrons in extra-nuclear part

Number of protons = 15

(e) Valency of element. The arrangement of electrons in orbits is 2, 8, 5. As the atom tends to gain three electrons, therefore it is trivalent electronegative (-3).

Question-9

15P31, 1H1, 18Ar40, 14Si30, 16S32, 19K40, 20Ca40, 1H2, 1H3.

 


Solution:
(a) 1H1, 1H2, 1H3 - isotopes (same number of atomic number)

(b) 18Ar40, 19K40, 20Ca40 - isobars (same number of mass number)

(c) 15P31, 14Si30, 16S32 - isotones (same number of neutrons)

Question-10

Which are isosters?

Solution:
Molecules having same numbers of atoms and also same number of electrons are called isosters.

Example: N2 and CO

N2 = 14 electrons

CO = 6 + 8 = 14 electrons.





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