I'm trying to make sense of this expansion. What is it and why. don't they do a mclaurin series expansion? If it is possible please do the Mclaurin expanision. If not just an explination. Thanks

icon
Related questions
Question
100%

I'm trying to make sense of this expansion. What is it and why. don't they do a mclaurin series expansion? If it is possible please do the Mclaurin expanision. If not just an explination. Thanks!

Because of how large n is, it's necessary to rewrite the expression in square brackets.
1
1
G²M²m³
2ħ²
[n² (1-1)² n²
ΔΕ Ξ
G²M²m³ 1
2ħ² n²
G² M²m
2ħ²n²
G² M²m³
2
3
(²¹ ²m³ [1 + (-2) (-) + (-²)X-2 -¹) (-1)² + (-²)(-²-1)(-2²-²2 (-²) + -]
(-2)(-2-1)
2ħ²n²
2!
3!
2ħ²n²
-2
[(₁ - 1)^²-1]
1--
n
2
3
4
----]
[2 ( ₁² ) + 3 ( ² ) ² + 4 (¹) ²³ + 5 ( ²1 ) *
1
www.stemjock.com
Therefore,
Zn²
Griffiths Quantum Mechanics 3e: Problem 4.20
ΔΕ =
The higher-order terms are extremely negligible compared to the first in the square brackets.
G² M²m² [2 (²)]
2ħ²n²
G²M²m
ħ²
GM
h.
= √√√ √²
+
(27) (1)
ħ 1
G²M²m² (Gm³ Mr.) (mg /GMr.)
(1
ħ²
≈(1.054571726 × 10-34 J·s)
(6.673 × 10-11 N.m²) (1.99 × 10³⁰ kg)
(1.496 x 1011 m)³
Page 6 of 6
AE≈ 2.10 x 10-4¹ J.
(3)
Transcribed Image Text:Because of how large n is, it's necessary to rewrite the expression in square brackets. 1 1 G²M²m³ 2ħ² [n² (1-1)² n² ΔΕ Ξ G²M²m³ 1 2ħ² n² G² M²m 2ħ²n² G² M²m³ 2 3 (²¹ ²m³ [1 + (-2) (-) + (-²)X-2 -¹) (-1)² + (-²)(-²-1)(-2²-²2 (-²) + -] (-2)(-2-1) 2ħ²n² 2! 3! 2ħ²n² -2 [(₁ - 1)^²-1] 1-- n 2 3 4 ----] [2 ( ₁² ) + 3 ( ² ) ² + 4 (¹) ²³ + 5 ( ²1 ) * 1 www.stemjock.com Therefore, Zn² Griffiths Quantum Mechanics 3e: Problem 4.20 ΔΕ = The higher-order terms are extremely negligible compared to the first in the square brackets. G² M²m² [2 (²)] 2ħ²n² G²M²m ħ² GM h. = √√√ √² + (27) (1) ħ 1 G²M²m² (Gm³ Mr.) (mg /GMr.) (1 ħ² ≈(1.054571726 × 10-34 J·s) (6.673 × 10-11 N.m²) (1.99 × 10³⁰ kg) (1.496 x 1011 m)³ Page 6 of 6 AE≈ 2.10 x 10-4¹ J. (3)
Expert Solution
Step 1

Given that the expression,we have

E=G2M2mE32h21n21-1n2-1n2=G2M2mE32h2n211-1n2-1E=G2M2mE32h2n21-1n-2-1----------(1)

Use the binomial expression,we have 

(1+x)n=1+nx1!+n(n-1)2!x2+n(n-1)(n-2)3!x3+n(n-1)(n-2)(n-3)4!x4........... here x<1

Compare  1-1n-2 with (1+x)n

Here,x=-1n,n=-2

Now, 

1-1n-2-1=1+-21!-1n+-2-2-12!-1n2+-2-2-1(-2-2)3!-1n3+-2-2-1(-2-2)(-2-3)4!-1n4..........-1                   =1+2n+62n2+246n3+255n4..........-1                   =1+2n+3n2+4n3+5n4.............-1

Put this value in the equation (1)

E=G2M2mE32h2n21-1n-2-1       =G2M2mE32h2n21+2n+3n2+4n3+5n4.............-1 E=G2M2mE32h2n221n+31n2+41n3+51n4.............

 

 

trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

Blurred answer
Similar questions