Question

# Can I please have a solution to this question. Young and Freedman University Physics With modern...

Can I please have a solution to this question. Young and Freedman University Physics With modern Physics 13th Edition. Chapter 41. Problem 43.

a) IDENTIFY and SET UP: The probability is P =? 2 dV with dV = 4? r2dr.
EXECUTE: ? 2 = A2e?2? r2 so P = 4? A2r2e?2? r2dr
(b) IDENTIFY and SET UP: P is maximum where dP 0.
dr
=
EXECUTE: d (r2e-2 ?r2 )= 0
dr
2re?2? r2 ? 4? r3e?2? r2 = 0 and this reduces to 2r ? 4? r3 = 0
r = 0 is a solution of the equation but corresponds to a minimum not a maximum. Seek r not equal to 0 so
divide by r and get 2 ? 4? r2 = 0.
This gives
r = 1/sqrt(2?)
= (We took the positive square root since r must be positive)

EVALUATE: This is different from the value of r, r = 0, where ? 2 is a maximum. At r = 0, ? 2
has a maximum but the volume element dV = 4? r2dr is zero here so P does not have a maximum
at r = 0.

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