4604_Solutions_Set1_fa07

4604_Solutions_Set1_fa07 - PHY4604 Fall 2007 Problem Set 1...

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PHY4604 Fall 2007 Problem Set 1 Solutions Department of Physics Page 1 of 8 PHY 4604 Problem Set #1 Solutions Problem 1 (12 points): Consider a room containing 14 people, whose ages are as follows: One person aged 14, One person aged 15, Three people aged 16, Two people aged 22, Two people aged 24, Five people aged 25. (a) (1 point) If you selected one person from the room, what is the probability that the person’s age would be 15? Answer: 1 in 14 . There are 14 people, all equally likely to be selected, of whom only one has age 15. (b) (1 point) What is the most probable age? Answer: 25 since more people have age 25. (c) (1 point) What is the median age? Answer: 23 , since 7 people have age greater than 23 and 7 have age less than 23. (d) (1 point) What is the average age? Answer: 21 Solution: 21 14 294 14 ) 25 ( 5 ) 24 ( 2 ) 22 ( 2 ) 16 ( 3 ) 15 ( ) 14 ( = = + + + + + (e) (1 point) Let N(j) be the number of people with age j. Histogram N(j) versus j. Distribution of Ages 0 1 2 3 4 5 6 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 j N(j) (f) (2 points) Compute <j 2 > and <j> 2 . Answer: <j 2 > = 459.57, <j> 2 = 441 Solution: 57 . 459 14 6434 14 ) 625 ( 5 ) 576 ( 2 ) 484 ( 2 ) 256 ( 3 ) 225 ( ) 196 ( 2 = = + + + + + >= < j (g) (3 points) Determine Δ j = j - <j> for each j, and compute the variance of the distribution using, σ 2 = <( Δ j) 2 >. What is the standard deviation, σ , of this distribution. Answer: σ 2 = 18.57, σ = 4.31 Solution: j=14: Δ j = 14 – 21 = -7, ( Δ j) 2 = 49 j=15: Δ j = 15 – 21 = -6, ( Δ j) 2 = 36 j=16: Δ j = 16 – 21 = -5, ( Δ j) 2 = 25
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PHY4604 Fall 2007 Problem Set 1 Solutions Department of Physics Page 2 of 8 j=22: Δ j = 22 – 21 = +1, ( Δ j) 2 = 1 j=24: Δ j = 24 – 21 = +3, ( Δ j) 2 = 9 j=25: Δ j = 25 – 21 = +4, ( Δ j) 2 = 16 57 . 18 14 260 14 ) 16 ( 5 ) 9 ( 2 ) 1 ( 2 ) 25 ( 3 ) 36 ( ) 49 ( ) ( 2 2 = = + + + + + >= Δ =< j σ and σ = 4.31. (h) (2 points) Compute the standard deviation using, 2 2 > < > < = j j , and show that you get the same answer as in part (g) . Answer: σ = 4.31 Solution: 57 . 18 14 260 14 6174 6434 ) 21 ( 14 6434 2 2 2 2 = = = = > < > =< j j and σ = 4.31 which is the same as in (g). Problem 2 (10 points): Consider the ( Gaussian ) wave function 2 2 ) ( ) ( a x Ae x = λ ψ , where A , a , and λ are positive real constants. The probability density is defined by ) ( ) ( ) ( ) ( 2 x x x x ρ = . (a) (1 point) Find the value of A that normalizes this wave function such that 1 ) ( = +∞ dx x . Answer: 4 1 = π A Solution: 2 2 1 2 0 2 2 ) ( 2 2 ) ( 2 2 ) ( 2 2 2 A A dy e A dy e A dx e A dx x y y a x = Γ = = = = +∞ +∞ +∞ +∞ Hence, = 2 A and 4 1 = A . I made the substitution y = x-a and used = Γ = = +∞ +∞ 2 ) ( 2 2 2 1 0 2 2 dy e dy e y y . Also note that 2 1 2 ) ( 2 2 2 3 0 2 2 2 2 = Γ = = +∞ +∞ dy e y dy e y y y 2 2 2 5 0 4 4 4 3 2 ) ( 2 2 2 2 = Γ = = +∞ +∞ dy e y dy e y y y
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PHY4604 Fall 2007 Problem Set 1 Solutions Department of Physics Page 3 of 8 0 2 = +∞ dy e y y n λ for n odd (b) (2 points) Find <x> and <x 2 > for this wave function. Answer: a x >= < , 2 2 2 1 a x + >= < Solution: a aA dy e aA dy e a y A dx xe A dx x x x y y a x = = = + = = >= < +∞ +∞ +∞ +∞ π ρ 2 0 2 2 ) ( 2 2 2 2 2 ) ( ) ( 2 2 2 2 0 2 2 0 2 2 2 2 ) ( 2 2 2 2 2 1 2 2 2 ) ( ) ( 2 2 2 2 a A a A dy e A a dy e y A dy e a y A dx e x A dx x x x y y y a x + = + = + = + = = >= < + + +∞ +∞
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4604_Solutions_Set1_fa07 - PHY4604 Fall 2007 Problem Set 1...

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