5. Hence show that: OU (hv exp(-ẞhv)) hv exp(-2ẞhv) =-3Nhv + oß (1-exp(-ßhv)) (1-exp(-ẞhv)) Hint 1: you will need to use the quotient rule: yf'(x) xf'(y) f'(x) f'(y) (f'(y))² (f'(y))² Hint 2: the following identity will avoid you having to use the chain rule: d (1 − e√(x)) = − f'(x)e√(x) dx
5. Hence show that: OU (hv exp(-ẞhv)) hv exp(-2ẞhv) =-3Nhv + oß (1-exp(-ßhv)) (1-exp(-ẞhv)) Hint 1: you will need to use the quotient rule: yf'(x) xf'(y) f'(x) f'(y) (f'(y))² (f'(y))² Hint 2: the following identity will avoid you having to use the chain rule: d (1 − e√(x)) = − f'(x)e√(x) dx
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question

Transcribed Image Text:5. Hence show that:
OU
(hv exp(-ẞhv))
hv exp(-2ẞhv)
=-3Nhv
+
oß
(1-exp(-ßhv)) (1-exp(-ẞhv))
Hint 1: you will need to use the quotient rule:
yf'(x) xf'(y)
f'(x)
f'(y) (f'(y))²
(f'(y))²
Hint 2: the following identity will avoid you having to use the chain rule:
d
(1 − e√(x)) = − f'(x)e√(x)
dx
Expert Solution
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