Atomic Structure

BITSAT 2015 Forum › Physical Chemistry questions › Atomic Structure

In a H-atom the energy in the nth orbit is given as E_n = -13.6n² ev . Show that E_n+1 - E_n = -13.6*2^n³ for large n values

How to solve??

#Chemistry #Physical Chemistry

UP 0 DOWN 0 0 2
Post on FacebookPost anonymously?

2 Answers

Debopam Purkayasthm ·2012-06-15 07:14:29

since n is infinitely large,n=n+1(approx)....again, 1/n is negligible....so take it 0...nd u will be able 2 solve d problm :)

UP 1 DOWN 2
Anirban Mondal HOW? PLEASE EXPLAIN THIS
Upvote·0· Reply ·2012-06-16 23:30:05
Anik Chatterjee couldnt get ,,....if n=n+1....En+1-En=En-En=0....!!
Upvote·0· Reply ·2012-06-17 02:44:28
Post on FacebookPost anonymously?

Debopam Purkayasthm ·2012-06-30 20:56:26

n approximtely equal to n+1....so the equatn will be -13.6[1/(n+1)² - 1/n²]= -13.6{-2n-1/n²(n+1)²}= 13.6{2n+1/n²(n+1)²}=13.6{2n/n²(n+1)²+1/n²(n+1)²}...now since n is infinitely lrge...n=n+1 approx...again...1/n⁴ is negligible.....so,the equatn stnds lyk 13.6{2n/n⁴ +1/n⁴},or,13.6{2/n³ +0}...which is equal to 13.6 *2/n³

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

Close [X]

Atomic Structure

2 Answers

Your Answer

Add Image

Similar Questions