Structure of Atom

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New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Matching Compounds to their Class:

o   Antacid: Cimetidine

o   Artificial Sweetener: Alitame

o   Antifertility: Novestrol

o   Tranquilizers: Valium

New question posted

a month ago

0 Follower 2 Views

New answer posted

a month ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

For n = 4, the possible values of l are 0, 1, 2, 3.
For l = 3, and m = -3.
Radial nodes = (n - l - 1) = (4 - 3 - 1) = 0
Ans = 0

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

ΔR? = (a? /3) (16 – 9)
ΔR? = (a? /2) (16 – 9)
∴ ΔR? /ΔR? = 2/3

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

For hydrogen atom :
For Lyman series n? = 1 & n? = ∞
1/λH = RH [1/1² - 1/∞²] So, λ = 1/RH
For He? ion Balmer series n? = 2 & n? = 3
1/λHe? = RH * Z² [1/n? ² - 1/n? ²]
1/λHe? = RH * 4 * [1/4 - 1/9] = RH * 4 * (5/36)
1/λHe? = (5/9)RH = (5/9) (1/λ)
(λHe? ) = (9/5)λ

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Balmer series lies in the visible region.

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2 months ago

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A
alok kumar singh

Contributor-Level 10

Charge / radius ratio of B e and A l is same because of diagonal relationship. Remaining statements are correct.

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

For H 2 O 2

Molarity =   Volume strength   11.2 = 5.6 11.2 = 0.5 M

Molarity = % ( w / w ) * 10 * d G M M

0.5 = % ( w / w ) * 10 * d 34

0.5 = % ( w / w ) * 10 * d 34

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