Structure of Atoms

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New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Configuration of the two elements are:

Al (Z = 13): [Ne]103s23p1

Si (Z = 14): [Ne]103s23p2
The unpaired electrons in silicon (Si) will experience more effective nuclear charge because the atomic number of the element Si is more than that of Al.

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Greater the penetration of the electron present in a particular orbital towards the nucleus more will be the magnitude of the effective nuclear charge. Based upon this

(i) 2s orbital is closer to the nucleus than 3s orbital and hence it will experience more effective nuclear charge.

(ii) 4d orbital will experience more effective nuclear charge due to its closer proximity to 4f orbital.

(iii) 3p orbital will experience more effective nuclear charge as it is closer to the nucleus

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3 months ago

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alok kumar singh

Contributor-Level 10

The nuclear charge experienced by electrons depends on the distance between the nucleus and orbital. The greater is this distance, the lesser is the effective nuclear charge. Among all the p orbitals, 4p orbital lies the farthest from the nucleus and thus experiences the lowest effective nuclear charge because of the maximum magnitude of screening or shielding effect.

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3 months ago

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alok kumar singh

Contributor-Level 10

The electrons may be assigned to the following orbitals :
(i) 4d
(ii) 3d
(iii) 4p
(iv) 3d
(v) 3p
(vi) 4p.
The increasing order of energy is :
(v) < (ii) = (iv) < (vi) = (iii) < (i)

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3 months ago

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alok kumar singh

Contributor-Level 10

Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of electron cannot be defined

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3 months ago

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alok kumar singh

Contributor-Level 10

ν = 4.37 x 105 m s-1, m = 0.1 kg

As per de Brogile's equation,

λ= m/v = (6.626 x 10-34 kg m2 s-1) / (0.1 kg) x (4.37 x 105 m s-1)

=6.626/0.437  x 10-34-5 m

= 1.516 x 10-38 m

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3 months ago

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New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

ν = 2.19 x 106 m s-1

As per de Brogile's equation,

λ= m/v = (6.626 x 10-34 kg m2 s-1) / (9.1 x 10-31 kg) x (2.19 x 106 m s-1)

6.626/91 * 2.19=  x 10-34+25 m = 0.33243 x 10-9 m = 332.43 pm

New answer posted

3 months ago

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alok kumar singh

Contributor-Level 10

λ = 800 pm = 800 x 10-12 m

m = 1.675 x 10-27 kg

As per de Brogile's equation,

ν =h/mλ  = (6.626 x 10-34 kg m2 s-1) / (1.675 x 10-27 kg) x (800 x 10-12 m)

=6.626/1.675 * 8  x 10-34+27+10

= 0.494 x 103 ms-1

= 494 ms-1

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

As per de Brogile's equation,

λ = h / mv = (6.626 x 10-34 kg m2 s-1) / (9.1 x 10-31 kg) x (1.6 x 106 ms-1)

= 0.455 x 10-9 m = 0.455 nm = 455 pm.

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