College Physics for AP® Courses

# Chapter 30

### Problems & Exercises

1.

$1 . 84 × 10 3 1 . 84 × 10 3 size 12{1 "." "84" times "10" rSup { size 8{3} } } {}$

3.

50 km

4.

$6 × 10 20 kg/m 3 6 × 10 20 kg/m 3 size 12{1 "." "93" times "10" rSup { size 8{"25"} } "kg/m" rSup { size 8{3} } } {}$

6.

(a) $10.0 μm 10.0 μm size 12{"10" "." 0" μm"} {}$

(b) It isn’t hard to make one of approximately this size. It would be harder to make it exactly $10.0 μm 10.0 μm size 12{"10" "." 0" μm"} {}$.

7.

$1λ=R1nf2−1ni2⇒λ=1R(ni⋅nf)2ni2−nf2;ni=2,nf=1,1λ=R1nf2−1ni2⇒λ=1R(ni⋅nf)2ni2−nf2;ni=2,nf=1, size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right ) drarrow λ= { {1} over {R} } left [ { { $$n rSub { size 8{i} } cdot n rSub { size 8{f} }$$ rSup { size 8{2} } } over {n rSub { size 8{i} } rSup { size 8{2} } - n rSub { size 8{f} } rSup { size 8{2} } } } right ]; n rSub { size 8{i} } =2" , "n rSub { size 8{f} } =1" ,"} {}$ so that

$λ = m 1.097 × 10 7 ( 2 × 1 ) 2 2 2 − 1 2 =1.22×10−7 m=122 nm λ = m 1.097 × 10 7 ( 2 × 1 ) 2 2 2 − 1 2 =1.22×10−7 m=122 nm size 12{λ= left ( { {m} over {1 "." "097" times "10" rSup { size 8{7} } } } right ) left [ { { $$2 "." 1$$ rSup { size 8{2} } } over {4 - 1} } right ]=1 "." "22" times "10" rSup { size 8{ - 7} } " m"="122"" nm"} {}$ , which is UV radiation.

9.

$a B = h 2 4π 2 m e kZq e 2 = ( 6.626 × 10 − 34 J·s ) 2 4π 2 ( 9.109 × 10 − 31 kg ) ( 8.988 × 10 9 N · m 2 / C 2 ) ( 1 ) ( 1.602 × 10 − 19 C ) 2 = 0.529 × 10 − 10 m a B = h 2 4π 2 m e kZq e 2 = ( 6.626 × 10 − 34 J·s ) 2 4π 2 ( 9.109 × 10 − 31 kg ) ( 8.988 × 10 9 N · m 2 / C 2 ) ( 1 ) ( 1.602 × 10 − 19 C ) 2 = 0.529 × 10 − 10 m$

11.

0.850 eV

13.

$2.12 × 10 –10 m 2.12 × 10 –10 m size 12{2 "." "12" times "10" rSup { size 8{"-10"} } " m"} {}$

15.

365 nm

It is in the ultraviolet.

17.

No overlap

365 nm

122 nm

19.

7

21.

(a) 2

(b) 54.4 eV

23.

$kZqe2rn2=meV2rn ,kZqe2rn2=meV2rn ,$ so that $rn=kZqe2meV2=kZqe2me1V2.rn=kZqe2meV2=kZqe2me1V2.$ From the equation $mevrn=nh2π ,mevrn=nh2π ,$ we can substitute for the velocity, giving: $rn=kZqe2me⋅4π2me2rn2n2h2rn=kZqe2me⋅4π2me2rn2n2h2$ so that $rn=n2Zh24π2mekqe2=n2ZaB ,rn=n2Zh24π2mekqe2=n2ZaB ,$ where $aB=h24π2mekqe2aB=h24π2mekqe2$.

25.

(a) $0.248× 10 −10 m0.248× 10 −10 m$

(b) 50.0 keV

(c) The photon energy is simply the applied voltage times the electron charge, so the value of the voltage in volts is the same as the value of the energy in electron volts.

27.

(a) $100×103eV100×103eV size 12{"100" times "10" rSup { size 8{3} } " eV"} {}$, $1.60×10−14J1.60×10−14J size 12{1 "." "60" times "10" rSup { size 8{ - "14"} } J} {}$

(b) $0.124×10−10 m0.124×10−10 m size 12{1 "." "24" times "10" rSup { size 8{ - "11"} } " m"} {}$

29.

(a) 8.00 keV

(b) 9.48 keV

30.

(a) 1.96 eV

(b) $( 1240 eV·nm ) / ( 1 . 96 eV ) = 633 nm ( 1240 eV·nm ) / ( 1 . 96 eV ) = 633 nm size 12{ $$"1240 eV·nm"$$ / $$1 "." "96 eV"$$ =" 633 nm"} {}$

(c) 60.0 nm

32.

693 nm

34.

(a) 590 nm

(b) $( 1240 eV·nm ) / ( 1 . 17 eV ) = 1.06 μm ( 1240 eV·nm ) / ( 1 . 17 eV ) = 1.06 μm size 12{ $$"1240 eV·nm"$$ / $$1 "." "96 eV"$$ =" 633 nm"} {}$

35.

$l=4, 3l=4, 3$ are possible since $l and $∣ml∣≤l∣ml∣≤l size 12{ lline m rSub { size 8{l} } rline {underline {<}} l} {}$.

37.

$n=4⇒l=3, 2, 1, 0⇒ml=±3,±2, ±1, 0n=4⇒l=3, 2, 1, 0⇒ml=±3,±2, ±1, 0$ are possible.

39.

(a) $1.49×10−34 J⋅s1.49×10−34 J⋅s size 12{1 "." "49" times "10" rSup { size 8{ - "34"} } " J" cdot s} {}$

(b) $1.06×10−34 J⋅s1.06×10−34 J⋅s size 12{1 "." "06" times "10" rSup { size 8{ - "34"} } " J" cdot s} {}$

41.

(a) $3.66×10−34 J⋅s3.66×10−34 J⋅s size 12{3 "." "66" times "10" rSup { size 8{ - "34"} } " J" cdot s} {}$

(b) $s=9.13×10−35 J⋅ss=9.13×10−35 J⋅s size 12{s=9 "." "14" times "10" rSup { size 8{ - "35"} } " J" cdot s} {}$

(c) $LS=123/4=4LS=123/4=4 size 12{ { {L} over {S} } = { { sqrt {"12"} } over { sqrt {3/4} } } =4} {}$

43.

$θ = 54.7º, 125.3º θ = 54.7º, 125.3º$

44.

(a) 32. (b) $2 in s, 6 in p, 10 in d,2 in s, 6 in p, 10 in d,$ and 14 in $ff size 12{f} {}$, for a total of 32.

46.

(a) 2

(b) $3d93d9 size 12{3d rSup { size 8{9} } } {}$

48.

(b) $n≥l n≥l$ is violated,

(c) cannot have 3 electrons in $s s$ subshell since $3 > ( 2 l + 1 ) = 2 3 > ( 2 l + 1 ) = 2$

(d) cannot have 7 electrons in $p p$ subshell since $7 > ( 2 l + 1 ) = 2 ( 2 + 1 ) = 6 7 > ( 2 l + 1 ) = 2 ( 2 + 1 ) = 6$

50.

(a) The number of different values of $mlml size 12{m rSub { size 8{l} } } {}$ is $±l,±(l−1),...,0±l,±(l−1),...,0$ for each $l>0l>0 size 12{l>0} {}$ and one for $l=0⇒(2l+1).l=0⇒(2l+1). size 12{l=0 drarrow $$2l+1$$ "." } {}$ Also an overall factor of 2 since each $mlml size 12{m rSub { size 8{l} } } {}$ can have $msms size 12{m rSub { size 8{s} } } {}$ equal to either $+1/2+1/2 size 12{+1/2} {}$ or $−1/2⇒2(2l+1)−1/2⇒2(2l+1) size 12{ - 1/2 drarrow 2 $$2l+1$$ } {}$.

(b) for each value of $ll size 12{l} {}$, you get $2(2l+1)2(2l+1) size 12{2 $$2l+1$$ } {}$

$=0, 1, 2, ...,(n–1)⇒2(2)(0)+1+(2)(1)+1+....+(2)(n−1)+1= 21+3+...+(2n−3)+(2n−1) ︸ n terms =0, 1, 2, ...,(n–1)⇒2(2)(0)+1+(2)(1)+1+....+(2)(n−1)+1= 21+3+...+(2n−3)+(2n−1) ︸ n terms size 12{ {}=0, 1," 2, " "." "." "." ", " $$"n–1"$$ drarrow 2 left lbrace left [ $$2$$ $$0$$ +1 right ]+ left [ $$2$$ $$1$$ +1 right ]+ "." "." "." "." + left [ $$2$$ $$n - 1$$ +1 right ] right rbrace = {2 left [1+3+ "." "." "." + $$2n - 3$$ + $$2n - 1$$ right ]} underbrace { size 8{n" terms"} } } {}$ to see that the expression in the box is $=n2,=n2,$ imagine taking $(n−1)(n−1) size 12{ $$n - 1$$ } {}$ from the last term and adding it to first term $=21+(n–1)+3+...+(2n−3)+(2n−1)–(n−1)=2n+3+....+(2n−3)+n.=21+(n–1)+3+...+(2n−3)+(2n−1)–(n−1)=2n+3+....+(2n−3)+n. size 12{ {}=2 left [1+ $$n"–1"$$ +3+ "." "." "." + $$2n - 3$$ + $$2n - 1$$ – $$n - 1$$ right ]=2 left [n+3+ "." "." "." "." + $$2n - 3$$ +n right ] "." } {}$ Now take $(n−3)(n−3) size 12{ $$n - 3$$ } {}$ from penultimate term and add to the second term $2 n+n+...+n+n ︸ n terms =2n22 n+n+...+n+n ︸ n terms =2n2 size 12{2 { left [n+n+ "." "." "." +n+n right ]} underbrace { size 8{n" terms"} } =2n rSup { size 8{2} } } {}$.

52.

The electric force on the electron is up (toward the positively charged plate). The magnetic force is down (by the RHR).

54.

401 nm

56.

(a) $6.54×10−16 kg6.54×10−16 kg size 12{6 "." "53" times "10" rSup { size 8{ - "16"} } " kg"} {}$

(b) $5.54×10−7 m5.54×10−7 m size 12{5 "." "53" times "10" rSup { size 8{ - 7} } " m"} {}$

58.

$1 . 76 × 10 11 C/kg 1 . 76 × 10 11 C/kg size 12{1 "." "76" times "10" rSup { size 8{"11"} } " C/kg"} {}$ , which agrees with the known value of $1 . 759 × 10 11 C/kg 1 . 759 × 10 11 C/kg size 12{1 "." "759" times "10" rSup { size 8{"11"} } " C/kg"} {}$ to within the precision of the measurement

60.

(a) 2.78 fm

(b) 0.37 of the nuclear radius.

62.

(a) $1.34×10231.34×1023 size 12{1 "." "34" times "10" rSup { size 8{"23"} } } {}$

(b) 2.52 MW

64.

(a) 6.42 eV

(b) $7.27×10−20 J/molecule7.27×10−20 J/molecule size 12{6 "." "75" times "10" rSup { size 8{ - "20"} } " J/molecule"} {}$

(c) 0.454 eV, 14.1 times less than a single UV photon. Therefore, each photon will evaporate approximately 14 molecules of tissue. This gives the surgeon a rather precise method of removing corneal tissue from the surface of the eye.

66.

91.18 nm to 91.22 nm

68.

(a) $1.24×1011 V1.24×1011 V size 12{1 "." "24" times "10" rSup { size 8{"11"} } " V"} {}$

(b) The voltage is extremely large compared with any practical value.

(c) The assumption of such a short wavelength by this method is unreasonable.

### Test Prep for AP® Courses

1.

(a), (d)

3.

(a)

5.

(a)

7.

(b)

9.

(a)

11.

(d)

13.

(d)

15.

(a), (c)

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