{"id":182410,"date":"2025-01-13T18:44:51","date_gmt":"2025-01-13T18:44:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182410"},"modified":"2025-01-13T18:44:54","modified_gmt":"2025-01-13T18:44:54","slug":"singly-ionized-carbon-atoms-are-accelerated-through-1-000-v-and-passed-intoa-mass-spectrometer-to-determine-the-isotopes-present","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/singly-ionized-carbon-atoms-are-accelerated-through-1-000-v-and-passed-intoa-mass-spectrometer-to-determine-the-isotopes-present\/","title":{"rendered":"Singly ionized carbon atoms are accelerated through 1 000 V and passed intoa mass spectrometer to determine the isotopes present"},"content":{"rendered":"\n

Singly ionized carbon atoms are accelerated through 1 000 V and passed intoa mass spectrometer to determine the isotopes present. The magnetic fieldstrength in the spectrometer is 0.200 T. (a) Determine the orbital radii for the12C and the 13C isotopes as they pass through the field. (b) Show that the ratioof the radii may be written in the formand verify that your radii in part (a) satisfy this formula.Answer: (a) r12=7.89cm r13=8.61cm (b) 0.9612. Radon gas has a half-life of 3.83 days. If 3.00 g of radon gas is present at timet =0, what mass of radon will remain after 1.50 days have passed?3. Ni6628(mass=65.9291u) undergoes beta decay to Cu6629(mass=65.9289u). (a) writethe complete decay formula for this process. (b) find the maximum kinetic energyof the emerging electrons. Answer (b) 186keV4. A by-product of some fission reactors is the isotope Pu23994, which is an alphaemitter with a half-life of 24000years: HeUPu4235923994??. Consider a sample of 1.0kg of pure Pu23994at t=0 and (b) the initial activity of thesample. (c) how long does the sample have to be stored if a safe activity level is0.10Bq? Answers5. the half-life of I131is 8.04 days. (a) convert the half-life to seconds. (b) calculatethe decay constant for this isotope. (c) convert Ci?500.0to the SI unit theBecquerel. (d) find the number of I131 nuclei necessary to produce a sample withan activity of Ci?50.0. (e) suppose the activity of certain I131sample is mCi40.6atgiven time. Find the number of half-lives the sample goes through in 40.2 decaysand the activity at the end of that period.6. A 200.0-mCi sample of a radioactive isotope is purchased by a medical supplyhouse. If the sample has a half-life of 14.0 days, how long will it keep before itsactivity is reduced to 20.0 mCi?7. After 2.00 days, the activity of a sample of an unknown type radioactivematerial has decreased to 84.2% of the initial activity. What is the half-life of thismaterial?<\/p>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Problem 1: Ionized Carbon Atoms in Mass Spectrometer<\/h3>\n\n\n\n

(a) Orbital Radii of C+\\text{C}^{+} Isotopes<\/h4>\n\n\n\n

To find the orbital radii of the singly ionized carbon isotopes 12C^{12}\\text{C} and 13C^{13}\\text{C} in a magnetic field, we can use the formula for the radius of the circular path of a charged particle in a magnetic field: r=mvqBr = \\frac{mv}{qB}<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • rr is the radius of the path.<\/li>\n\n\n\n
  • mm is the mass of the particle.<\/li>\n\n\n\n
  • vv is the velocity of the particle.<\/li>\n\n\n\n
  • qq is the charge of the particle.<\/li>\n\n\n\n
  • BB is the magnetic field strength.<\/li>\n<\/ul>\n\n\n\n

    First, we need to find the velocity of the particles. Since the carbon atoms are accelerated through a potential difference VV, the kinetic energy gained by the ion is: 12mv2=qV\\frac{1}{2}mv^2 = qV<\/p>\n\n\n\n

    Thus, the velocity vv is given by: v=2qVmv = \\sqrt{\\frac{2qV}{m}}<\/p>\n\n\n\n

    Now, we substitute this expression for vv into the formula for the radius rr: r=m2qVmqB=2mVqBr = \\frac{m \\sqrt{\\frac{2qV}{m}}}{qB} = \\frac{\\sqrt{2mV}}{qB}<\/p>\n\n\n\n

    For 12C^{12}\\text{C} and 13C^{13}\\text{C}, the only difference is the mass, so the radii will be proportional to the square root of the mass.<\/p>\n\n\n\n

    For 12C^{12}\\text{C}, the mass is 12\u2009u12 \\, \\text{u} and for 13C^{13}\\text{C}, the mass is 13\u2009u13 \\, \\text{u}. The charge qq for both is the same (since both are singly ionized), and the magnetic field strength B=0.200\u2009TB = 0.200 \\, \\text{T} and the potential V=1000\u2009VV = 1000 \\, \\text{V} are also the same. Using the given masses and the formula above, we calculate the orbital radii.<\/p>\n\n\n\n

    (b) Ratio of the Radii<\/h4>\n\n\n\n

    The ratio of the radii for 12C^{12}\\text{C} and 13C^{13}\\text{C} is given by: r13r12=m13m12\\frac{r_{13}}{r_{12}} = \\sqrt{\\frac{m_{13}}{m_{12}}}<\/p>\n\n\n\n

    Using the values of m13=13\u2009um_{13} = 13 \\, \\text{u} and m12=12\u2009um_{12} = 12 \\, \\text{u}, we get: r13r12=1312\u22481.061\\frac{r_{13}}{r_{12}} = \\sqrt{\\frac{13}{12}} \\approx 1.061<\/p>\n\n\n\n

    Thus, the ratio is approximately 1.061.<\/p>\n\n\n\n

    \n\n\n\n

    Problem 2: Radon Gas Half-Life<\/h3>\n\n\n\n

    The half-life t1\/2t_{1\/2} of radon gas is 3.83 days, and the initial mass m0m_0 is 3.00 g. We are asked to find the remaining mass after 1.50 days.<\/p>\n\n\n\n

    The formula for exponential decay is: m(t)=m0(12)tt1\/2m(t) = m_0 \\left( \\frac{1}{2} \\right)^{\\frac{t}{t_{1\/2}}}<\/p>\n\n\n\n

    Substituting the given values: m(1.50)=3.00(12)1.503.83\u22483.00\u00d70.798=2.39\u2009gm(1.50) = 3.00 \\left( \\frac{1}{2} \\right)^{\\frac{1.50}{3.83}} \\approx 3.00 \\times 0.798 = 2.39 \\, \\text{g}<\/p>\n\n\n\n

    So, after 1.50 days, approximately 2.39 g of radon will remain.<\/p>\n\n\n\n

    \n\n\n\n

    Problem 3: Beta Decay of Ni66\\text{Ni}^{66}<\/h3>\n\n\n\n

    (a) Complete Decay Formula<\/h4>\n\n\n\n

    The beta decay process for Ni66\\text{Ni}^{66} to Cu66\\text{Cu}^{66} can be written as: Ni2866\u2192Cu2966+\u03b2\u2212+\u03bd\u203ee\\text{Ni}^{66}_{28} \\rightarrow \\text{Cu}^{66}_{29} + \\beta^{-} + \\overline{\\nu}_e<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • \u03b2\u2212\\beta^{-} represents the emitted electron.<\/li>\n\n\n\n
    • \u03bd\u203ee\\overline{\\nu}_e is the electron antineutrino.<\/li>\n<\/ul>\n\n\n\n

      (b) Maximum Kinetic Energy of the Electron<\/h4>\n\n\n\n

      The maximum kinetic energy of the emitted electron is given by the difference in mass between the parent and daughter nuclei. The energy released in the decay is: E=(Mass of Ni66\u2212Mass of Cu66)c2E = (\\text{Mass of Ni}^{66} – \\text{Mass of Cu}^{66}) c^2<\/p>\n\n\n\n

      Given that the mass of Ni66=65.9291\u2009u\\text{Ni}^{66} = 65.9291 \\, \\text{u} and Cu66=65.9289\u2009u\\text{Cu}^{66} = 65.9289 \\, \\text{u}, the mass difference is: \u0394m=65.9291\u221265.9289=0.0002\u2009u\\Delta m = 65.9291 – 65.9289 = 0.0002 \\, \\text{u}<\/p>\n\n\n\n

      Converting this to energy: E=0.0002\u00d7931.5\u2009MeV=0.186\u2009MeV=186\u2009keVE = 0.0002 \\times 931.5 \\, \\text{MeV} = 0.186 \\, \\text{MeV} = 186 \\, \\text{keV}<\/p>\n\n\n\n

      Thus, the maximum kinetic energy of the emerging electron is 186 keV.<\/p>\n\n\n\n

      \n\n\n\n

      Problem 4: Pu-239 Alpha Decay<\/h3>\n\n\n\n

      (a) Alpha Decay Equation<\/h4>\n\n\n\n

      The alpha decay of Pu239\\text{Pu}^{239} can be written as: Pu94239\u2192U92235+\u03b1\\text{Pu}^{239}_{94} \\rightarrow \\text{U}^{235}_{92} + \\alpha<\/p>\n\n\n\n

      Where \u03b1\\alpha represents the emitted alpha particle (helium nucleus).<\/p>\n\n\n\n

      (b) Initial Activity of the Sample<\/h4>\n\n\n\n

      The initial activity A0A_0 is given by: A0=\u03bbNA_0 = \\lambda N<\/p>\n\n\n\n

      Where \u03bb\\lambda is the decay constant and NN is the number of atoms. For a sample of 1.0 kg of Pu239\\text{Pu}^{239}, we first calculate the number of atoms using the molar mass: N=mM\u00d7NAN = \\frac{m}{M} \\times N_A<\/p>\n\n\n\n

      Where m=1.0\u2009kgm = 1.0 \\, \\text{kg}, M=239\u2009g\/molM = 239 \\, \\text{g\/mol}, and NA=6.022\u00d71023\u2009atoms\/molN_A = 6.022 \\times 10^{23} \\, \\text{atoms\/mol}.<\/p>\n\n\n\n

      The decay constant \u03bb\\lambda is related to the half-life by: \u03bb=ln\u20612t1\/2\\lambda = \\frac{\\ln 2}{t_{1\/2}}<\/p>\n\n\n\n

      Using the half-life of Pu239\\text{Pu}^{239}, which is 24,000 years, we calculate the activity.<\/p>\n\n\n\n

      (c) Safe Activity Level<\/h4>\n\n\n\n

      We can calculate the time required for the activity to drop to a safe level using the decay law: A(t)=A0(12)tt1\/2A(t) = A_0 \\left( \\frac{1}{2} \\right)^{\\frac{t}{t_{1\/2}}}<\/p>\n\n\n\n

      We solve for tt when A(t)=0.10\u2009BqA(t) = 0.10 \\, \\text{Bq}.<\/p>\n\n\n\n

      \n\n\n\n

      Problem 5: Iodine-131<\/h3>\n\n\n\n

      (a) Half-Life Conversion<\/h4>\n\n\n\n

      The half-life of iodine-131 is given as 8.04 days. To convert this to seconds: t1\/2=8.04\u00d724\u00d73600=693,024\u2009secondst_{1\/2} = 8.04 \\times 24 \\times 3600 = 693,024 \\, \\text{seconds}<\/p>\n\n\n\n

      (b) Decay Constant<\/h4>\n\n\n\n

      The decay constant \u03bb\\lambda is: \u03bb=ln\u20612t1\/2=ln\u20612693,024\u22481.00\u00d710\u22126\u2009s\u22121\\lambda = \\frac{\\ln 2}{t_{1\/2}} = \\frac{\\ln 2}{693,024} \\approx 1.00 \\times 10^{-6} \\, \\text{s}^{-1}<\/p>\n\n\n\n

      (c) Conversion to Becquerels<\/h4>\n\n\n\n

      To convert 500.0 Ci to Becquerels: 1\u2009Ci=3.7\u00d71010\u2009Bq1 \\, \\text{Ci} = 3.7 \\times 10^{10} \\, \\text{Bq}<\/p>\n\n\n\n

      Thus: 500.0\u2009Ci=500.0\u00d73.7\u00d71010=1.85\u00d71013\u2009Bq500.0 \\, \\text{Ci} = 500.0 \\times 3.7 \\times 10^{10} = 1.85 \\times 10^{13} \\, \\text{Bq}<\/p>\n\n\n\n

      (d) Number of Nuclei for Activity<\/h4>\n\n\n\n

      Using the equation A=\u03bbNA = \\lambda N, we solve for NN given the activity and decay constant.<\/p>\n\n\n\n

      (e) Number of Half-Lives for Activity Decay<\/h4>\n\n\n\n

      Using the decay equation, we can find how many half-lives correspond to a given number of decays.<\/p>\n\n\n\n

      \n\n\n\n

      Problem 6: Half-Life of Radioactive Isotope<\/h3>\n\n\n\n

      To find how long it will take for the activity of a 200.0 mCi sample to decrease to 20.0 mCi, we use the formula for exponential decay: A(t)=A0(12)tt1\/2A(t) = A_0 \\left( \\frac{1}{2} \\right)^{\\frac{t}{t_{1\/2}}}<\/p>\n\n\n\n

      By solving for tt, we find the time needed for the activity to drop.<\/p>\n\n\n\n

      \n","protected":false},"excerpt":{"rendered":"

      Singly ionized carbon atoms are accelerated through 1 000 V and passed intoa mass spectrometer to determine the isotopes present. The magnetic fieldstrength in the spectrometer is 0.200 T. (a) Determine the orbital radii for the12C and the 13C isotopes as they pass through the field. (b) Show that the ratioof the radii may be […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-182410","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182410","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=182410"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182410\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182410"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182410"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182410"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}