(a) 0.3 c. (b) 0.7 c. (c) 1.0 c. (d) 1.7 c.

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1 Seat: PHYS 1510 (Spring 2013) Exam #4, V1 Name: Multiple choice (4 pts each) circle the correct answers. 1. Three isotopes have atomic number 10: beryllium, boron, and carbon. The atom 10 4Be has mass u, the atom 10 5B has mass u, and the atom 10 6C has mass u. Two of them decay by emitting beta rays. Which one of them is stable? (a) The Be isotope is stable. (b) The B isotope is stable. (c) The C isotope is stable. (d) There is insufficient information to know which is stable. 2. Based on book CQ27.5 A spaceship is traveling directly toward you at a speed of 0.7 c. It fires a laser beam at you. You measure the speed of the beam to be (a) 0.3 c. (b) 0.7 c. (c) 1.0 c. (d) 1.7 c. 3. Based on book Prob You have the option of measuring the position of an electron with an accuracy of 0.1 nm or 0.2 nm. Which accuracy will allow the most accurate measurement of the speed? (a) 0.1 nm (b) 0.2 nm (c) Either one will work. 4. From book Prob 29.5 How many neutrons and electrons are contained in the ion 9 4Be +? [Choices below are (number of neutrons, number of electrons)] (a) (9,4) (b) (9,3) (c) (4,9) (d) (13,9) (e) (5,4) (f) none of these 5. From book Prob simplified The allowed energies of a simple atom are 0.0 ev, 5.0 ev, and 8.0 ev. An electron with energy 7.0 ev collides with the atom. What is the least energy the electron can have after the collision? (a) 0.0 ev (b) 1.0 ev (c) 2.0 ev (d) 5.0 ev (e) 7.0 ev (f) 8.0 ev 6. From book MC Which of the following decays are possible? (a) U Th + α (b) U Th + γ (c) U Th + e (d) None are possible. (e) All are possible.

2 Problems: Show all work to receive full credit. In all problems choose the positive x-direction to be to the right and the positive y-direction to be up. 6 pts 1. From book Prob simplified You start with radioactive nuclei with a half-life of 10.0 minutes. (a) What is the initial activity? (b) How long does it take for the activity to equal that value? The activity is R = N/τ with τ = t 1/2 / ln 2. You can use dec/sec or Bq for the unit of activity. To find the time to reach the fraction, use frac = e t/τ. Use the ln to undo the exponential: ln(frac) = t/τ. This means the time is t = τ ln(frac). τ = 600 s ln = 866 s R = = Bq 866 s t = 866 s ln(0.167) = 1550 s (V 1) τ = 480 s ln = 692 s R = = Bq 692 s t = 692 s ln(0.172) = 1220 s (V 2) τ = 360 s ln = 519 s R = = Bq 519 s t = 519 s ln(0.159) = 955 s (V 3) 6 pts 2. Based on book Prob The evil space lord Grtlxcfzr is attacking you with the feared Auoioua rocket which fires at a speed of c relative to the ship. Grtlxcfzr s ship approaches you at a speed of c. According to your crew, with what speed is the Auoioua rocket approaching? This is a problem where you have to add velocities relativistically. In this problem, u is the velocity of the rocket and v is the velocity of the ship. Both are positive. You need to used u = (u + v)/(1 + u v/c 2 ) u = u = u = 0.4 c c 1 + ( ) = c = 0.75 c (V 1) 0.55 c c 1 + ( ) = 0.85 c = 0.73 c (V 2) c c 1 + ( ) = 0.81 c = 0.72 c (V 3) 1.126

3 6 pts 3. Based on book Prob simplified A hydrogen atom has the energy of ev. What is the diameter of the electron s orbit? You can find the diamater by multiplying the radius by 2: d = 2n 2 a B. To find n 2, you can use the expression for the energy: E = 13.6 ev/n 2. This means n 2 = 13.6 ev/e. n 2 = n 2 = 13.6 ev 0.85 ev 13.6 ev ev = 16 d = nm = 1.70 nm (V 1) = 25 d = nm = 2.65 nm (V 2) n 2 = 13.6 ev 3.4 ev = 4 d = nm = nm (V 3) 6 pts 4. From book Prob 28.11, You shine light with a wavelength of nm on a material. Electrons come out of the material with a speed as high as m/s. What is the work function of the material? The work function is how much energy is needed to remove an electron from the material. Conservation of energy gives K max = hf E 0. Use the property of light that f = c/λ to get the energy of the photon. You don t have to convert to ev. K max = kg( m/s) 2 = J E ph = J s m/s m = J E 0 = J J = J = 1.46 ev (V 1) K max = kg( m/s) 2 = J E ph = J s m/s m = J E 0 = J J = J = 1.90 ev (V 2) K max = kg( m/s) 2 = J E ph = J s m/s m = J E 0 = J J = J = 2.15 ev (V 3)

4 12 pts 5. From book Probs and An electron starts at rest in a potential of V. A short time later it reaches a potential of V after which the potential is constant. It then travels through a tube that is 3.30 m long. (a) What is the electron s final speed? (b) According to the electron, how long does it take to travel through the tube? You need to use conservation of energy to find the final kinetic energy: K f + qv f = K i + V i. The electron starts at rest and the charge q = e. This gives K f = e(v i V f ) = e(v f V i ). If you use the nonrelativistic expression, K = (1/2)mv 2, you will get a speed larger than c. So you need to use relativity. Add the rest energy to K to get the total energy: E = γmc 2 = mc 2 + K. Divide by mc 2 to get γ = 1/ 1 u 2 /c 2 = 1 + (K/mc 2 ). For electrons mc 2 = J. Solve for u. For (b), you need to length contract the tube L = l 1 β 2 and divide that by the speed to get the time in the tube according to the electron. K f = C V = J γ = J J = u2 c 2 = = u 2 c 2 = L = 3.3 m = m t = u = c m m/s = 5.71 ns (V 1) K f = C V = J γ = J J = u2 c 2 = = u 2 c 2 = L = 2.7 m = m t = u = c m m/s = 4.20 ns (V 2) K f = C V = J γ = J J = u2 c 2 = = u 2 c 2 = L = 2.2 m = m t = u = c m m/s = 3.11 ns (V 3)

5 Equations Basic Mathematic Formulas sin θ = O/H cos θ = A/H tan θ = O/A H 2 = O 2 + A 2 A circ = πr 2 C circ = 2πr V sph = 4π 3 r3 A sur of sph = 4πr 2 x = b ± b 2 4ac 2a F net = F 1 + F = m a τ = rf = r F = rf sin ϕ F = mv 2 /r m p = kg m e = kg 1 u = kg m α = 4u index of refraction: Water (1.33), Ethyl Alc. (1.36), Glass (1.50), Diamond (2.42) ( [ x y(x, t) = A cos 2π λ ± t ]) T Chapter 15 T = 1/f v = fλ v str = T s /µ v snd = 343 m/s c = m/s I = P/A I = P src /(4πr 2 ) f = f 0 /(1±[v s /v]) f ± = (1±[v o /v])f 0 Chapter 16 λ m = 2L/m f m = m(v/[2l]) m = 1, 2, 3... Chapter 17 w = 2λL/a a sin θ p = pλ y p = L tan θ p p = 1, 2, 3... d sin θ m = mλ d sin θ m = (m + 1/2)λ y m = L tan θ m m = 0, 1, 2... Chapter 18 θ r = θ i n 1 sin θ 1 = n 2 sin θ 2 θ c = sin 1 (n 2 /n 1 ) v = c n 1 s + 1 s = 1 f m = s s Chapter 19 d min = 0.61λ/NA θ 1 = 1.22λ/D P = 1/f M = θ/θ 0 = 25 cm/f M = f o /f e M = L 25 cm/(f o f e ) Chapter 20 F 1 on 2 = K q 1 q 2 /r 2 K = N m 2 /C 2 e = C E = Fon q /q E tot = E 1 + E E = K q /r 2 E = Q/(ε 0 A) ε 0 = C 2 /(N m 2 ) Chapter 21 U elec = qv K f + qv f = K i + qv i V = Kq/r V C = Q/C C = κε 0 A/d U C = 1 2 C( V c) 2 u E = 1 2 κε 0E 2 V = x d V c E = V C d 1 ev = J

6 Chapter 22 I = q/ t R = ρl/a I = V/R P emf = IE P R = I V R = I 2 R = ( V R ) 2 /R V i = 0 i Chapter 23 Iin = I out R eq = R 1 +R (1/C eq ) = (1/C 1 )+(1/C 2 )+... (1/R eq ) = (1/R 1 ) + (1/R 2 ) +... C eq = C 1 + C Chapter 24 F = qvb sin α F = ILB τ = (IA)B sin θ r = mv/( q B) B = µ 0 I/(2πr) B = µ 0 I/(2R) B = µ 0 IN/L µ 0 = T m/a Chapter 25 Φ = AB cos θ E = Φ/ t E 0 = cb 0 E = hf I = I 0 cos 2 θ λ peak = nm K/T β = v/c Chapter 27 t = τ/ 1 β 2 L = 1 β 2 l p = γmu γ = 1/ 1 u 2 /c 2 E = γmc 2 = mc 2 +(γ 1)mc 2 u = (u v)/(1 uv/c 2 ) u = (u +v)/(1+u v/c 2 ) Chapter 28 h = J s λ = h/(mv) x p x h/(2π) V stop = K max /e = (hf E 0 )/e 2d cos θ = mλ m = 1, 2, 3,... Chapter 29 E phot = E atom = E f E i 2πr = nλ E n = 13.6 ev/n 2 r n = n 2 a B a B = nm λ n m = 91.1 nm/(1/m 2 1/n 2 ) m = 1, 2, 3... n = m + 1, m + 2, m h = h/(2π) L = l(l + 1) h l = 0, 1, 2... m = l, l+1, , l 1, l m s = ±1/2 A = Z + N Chapter 30 B = (Zm H + Nm n m atom ) MeV/u A ZX A 4 Z 2 Y + α + E A ZX A Z+1 Y +β+e N = N 0 e t/τ N = N 0 (1/2) t/t 1/2 t 1/2 = τ ln 2 R = N/τ dose equiv (Sv) = dose(gy) RBE