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GRADE 11 PRACTICE

Physics Problems: Nuclear physics, Grade 11

Nuclear composition, binding energy and radioactivity.

20 problems across 2 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

20 problems
Position in this set: 1 from 20

Problem #76

Nuclear physics

What is the minimum distance an alpha particle can approach a gold nucleus with a speed of 2\cdot10^{7}\,\text{м/с}? Charge number of gold Z = 79, alpha particle mass 6.68·10⁻²⁷ kg, electron charge modulus 1.6·10⁻¹⁹ Cl. Consider the gold core motionless, a head-on collision.

Answer options

  1. 2{,}72 \cdot 10^{-13}\,\text{м}
  2. 2{,}72 \cdot 10^{-14}\,\text{м}
  3. 1{,}36 \cdot 10^{-14}\,\text{м}
  4. 5{,}44 \cdot 10^{-14}\,\text{м}
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Given
v
2\cdot 10^{7}\,\text{м/с}
q_1
2\cdot 1{,}6\cdot 10^{-19}\,\text{Кл}
q_2
79\cdot 1{,}6\cdot 10^{-19}\,\text{Кл}
m_1
4\cdot 1{,}67\cdot 10^{-27}\,\text{кг}
Find

r

  1. \dfrac{m_1 v^{2}}{2} = \dfrac{kq_1 q_2}{r}

    At the point of closest approach, all kinetic energy has turned into potential energy.

  2. r = \dfrac{2kq_1 q_2}{m_1 v^{2}}r = \dfrac{2\cdot 9\cdot 10^{9}\cdot 2\cdot 1{,}6\cdot 10^{-19}\cdot 79\cdot 1{,}6\cdot 10^{-19}}{4\cdot 1{,}67\cdot 10^{-27}\cdot\left(2\cdot 10^{7}\right)^{2}}r = 2{,}72\cdot 10^{-14}\,\text{м}

Answerr = 2{,}72\cdot 10^{-14}\,\text{м}

Position in this set: 2 from 20

Problem #78

Nuclear physics

During the alpha decay of a resting uranium-238 nucleus, an alpha particle is emitted at a speed 2\cdot10^{7}\,\text{м/с}. What speed does the child core get? Compare its kinetic energy with the energy of an alpha particle.

Answer options

  1. \left(1{,}7 \cdot 10^{5}\,\text{м/с};\ 0{,}00855\,\text{доля энергии}\right)
  2. \left(6{,}8 \cdot 10^{5}\,\text{м/с};\ 0{,}0342\,\text{доля энергии}\right)
  3. \left(3{,}4 \cdot 10^{6}\,\text{м/с};\ 0{,}171\,\text{доля энергии}\right)
  4. \left(3{,}4 \cdot 10^{5}\,\text{м/с};\ 0{,}0171\,\text{доля энергии}\right)
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Given
m_1
238\,\text{а.е.м.}
m_2
4\,\text{а.е.м.}
v_2
2\cdot 10^{7}\,\text{м/с}
Find

v_1\dfrac{E_1}{E_2}

  1. \left(m_1 - m_2\right)v_1 = m_2 v_2

    The nucleus was at rest, so the momenta of the fragment and the alpha particle are equal in magnitude.

  2. v_1 = \dfrac{m_2 v_2}{m_1 - m_2}v_1 = \dfrac{4\cdot 2\cdot 10^{7}}{238 - 4}v_1 = 3{,}4\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}
  3. \dfrac{E_1}{E_2} = \dfrac{\left(m_1 - m_2\right)v_1^{2}}{m_2 v_2^{2}} = \dfrac{m_2}{m_1 - m_2}\dfrac{E_1}{E_2} = \dfrac{4}{234}\dfrac{E_1}{E_2} = 0{,}0171

    The recoil core gets less than two percent of the energy - almost all of it is carried away by the alpha particle.

Answerv_1 = 3{,}4\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}\dfrac{E_1}{E_2} = 0{,}0171

Position in this set: 3 from 20

Problem #79

Nuclear physics

The uranium-238 nucleus, as a result of alpha and beta decays, turned into a lead-206 nucleus. Determine the number of alpha and beta decays.

Answer options

  1. \left(8\,\alpha\text{-распадов};\ 6\,\beta\text{-распадов}\right)
  2. \left(4\,\alpha\text{-распадов};\ 3\,\beta\text{-распадов}\right)
  3. \left(16\,\alpha\text{-распадов};\ 12\,\beta\text{-распадов}\right)
  4. \left(80\,\alpha\text{-распадов};\ 60\,\beta\text{-распадов}\right)
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Given
\text{reaction}
^{238}_{92}\mathrm{U} \to {}^{206}_{82}\mathrm{Pb} + x\,{}^{4}_{2}\mathrm{He} + y\,{}^{0}_{-1}e
Find

xy

  1. \left|\Delta A\right| = 238 - 206 = 32\ \Rightarrow\ x = \dfrac{32}{4}x = 8

    The mass number is changed only by alpha particles, each carrying 4.

  2. q_{\alpha} = 2x = 16;\quad q_e = 92 - 82 - 16q_e = -6
  3. y = \dfrac{q_e}{-1}y = 6

    Result for balancing A and Z: ²³⁸₉₂U → ²⁰⁶₈₂Pb + 8·⁴₂He + 6·⁰₋₁e. Antineutrinos are not shown here.

Answerx = 8y = 6

Position in this set: 4 from 20

Problem #80

Nuclear physics

Available 8\,\text{кг} radioactive cesium-137. To calculate, round the half-life to 30 years Find the mass of cesium that decayed during 150 years

Answer options

  1. 77{,}5\,\text{кг}
  2. 7{,}75\,\text{кг}
  3. 3{,}875\,\text{кг}
  4. 15{,}5\,\text{кг}
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Given
m_0
8\,\text{кг}
T
30\,\text{лет}
t
150\,\text{лет}
Find

m_0 - m

  1. \dfrac{t}{T}=\dfrac{150}{30}\dfrac{t}{T} = 5

    During the observation period, exactly five half-lives passed.

  2. m = m_0\cdot 2^{-t/T}m = 8\cdot 2^{-5}m = 0{,}25\,\text{кг}
  3. m_0 - m = m_0\left(1 - 2^{-t/T}\right)8\left(1 - \dfrac{1}{32}\right) = 8\cdot\dfrac{31}{32}m_0 - m = 7{,}75\,\text{кг}

Answerm_0 - m = 7{,}75\,\text{кг}

Position in this set: 5 from 20

Problem #81

Nuclear physics

How much energy will be released if the mass of the system decreases by 1\,\text{а.е.м.}?

Answer options

  1. 2{,}98 \cdot 10^{-10}\,\text{Дж}
  2. 1{,}49 \cdot 10^{-9}\,\text{Дж}
  3. 1{,}49 \cdot 10^{-10}\,\text{Дж}
  4. 7{,}45 \cdot 10^{-11}\,\text{Дж}
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Given
\Delta m
1\,\text{а.е.м.}1{,}66\cdot 10^{-27}\,\text{кг}
c
3\cdot 10^{8}\,\text{м/с}
Find

E

  1. E = \Delta m\,c^{2}1{,}66\cdot 10^{-27}\cdot\left(3\cdot 10^{8}\right)^{2}E = 1{,}49\cdot 10^{-10}\,\text{Дж} = 931\,\text{МэВ}

    This is where the conversion factor of 931 MeV per atomic mass unit is taken.

AnswerE = 1{,}49\cdot 10^{-10}\,\text{Дж}

Position in this set: 6 from 20

Problem #82

Nuclear physics

The mass of a beryllium-9 atom is 9{,}01219\,\text{а.е.м.}. Determine the mass defect and binding energy of the nucleus. For calculation, take the masses of the proton, neutron and electron equal to 1.00728; 1.00866 and 0.00055 amu; 1 amu·c² = 931 MeV.

Answer options

  1. \left(0{,}031215\,\text{а.е.м.};\ 29{,}05\,\text{МэВ}\right)
  2. \left(0{,}12486\,\text{а.е.м.};\ 116{,}2\,\text{МэВ}\right)
  3. \left(0{,}6243\,\text{а.е.м.};\ 581\,\text{МэВ}\right)
  4. \left(0{,}06243\,\text{а.е.м.};\ 58{,}1\,\text{МэВ}\right)
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Given
m_a
9{,}01219\,\text{а.е.м.}
Z
4
N
5
m_p
1{,}00728\,\text{а.е.м.}
m_n
1{,}00866\,\text{а.е.м.}
m_e
0{,}00055\,\text{а.е.м.}
Find

\Delta mE_{\text{св}}

  1. m_{\text{я}} = m_a - Zm_em_{\text{я}} = 9{,}01219 - 4\cdot 0{,}00055m_{\text{я}} = 9{,}00999\,\text{а.е.м.}

    We subtract the mass of the electron shell from the mass of the atom.

  2. \Delta m = Zm_p + Nm_n - m_{\text{я}}\Delta m = 4\cdot 1{,}00728 + 5\cdot 1{,}00866 - 9{,}00999\Delta m = 0{,}06243\,\text{а.е.м.}
  3. E_{\text{св}} = \Delta m\cdot 931E_{\text{св}} = 0{,}06243\cdot 931E_{\text{св}} = 58{,}1\,\text{МэВ}

Answer\Delta m = 0{,}06243\,\text{а.е.м.}E_{\text{св}} = 58{,}1\,\text{МэВ}

Position in this set: 7 from 20

Problem #83

Nuclear physics

Determine the mass defect, total and specific binding energy of the helium-4 nucleus. The mass of a helium-4 atom is 4.0026 amu. For calculation, take the masses of the proton, neutron and electron equal to 1.00728; 1.00866 and 0.00055 amu; 1 amu·c² = 931 MeV.

Answer options

  1. \left(0{,}03038\,\text{а.е.м.};\ 28{,}28\,\text{МэВ};\ 7{,}07\,\text{МэВ/нуклон}\right)
  2. \left(0{,}01519\,\text{а.е.м.};\ 14{,}14\,\text{МэВ};\ 3{,}535\,\text{МэВ/нуклон}\right)
  3. \left(0{,}06076\,\text{а.е.м.};\ 56{,}56\,\text{МэВ};\ 14{,}14\,\text{МэВ/нуклон}\right)
  4. \left(0{,}3038\,\text{а.е.м.};\ 282{,}8\,\text{МэВ};\ 70{,}7\,\text{МэВ/нуклон}\right)
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Given
\text{core}
^{4}_{2}\mathrm{He}
m_a
4{,}0026\,\text{а.е.м.}
Z
2
N
2
m_p
1{,}00728\,\text{а.е.м.}
m_n
1{,}00866\,\text{а.е.м.}
m_e
0{,}00055\,\text{а.е.м.}
Find

\Delta mE_{\text{св}}E'_{\text{св}}

  1. m_{\text{я}} = m_a - Zm_em_{\text{я}} = 4{,}0026 - 2\cdot 0{,}00055m_{\text{я}} = 4{,}0015\,\text{а.е.м.}
  2. \Delta m = Zm_p + Nm_n - m_{\text{я}}\Delta m = 2\cdot 1{,}00728 + 2\cdot 1{,}00866 - 4{,}0015\Delta m = 0{,}03038\,\text{а.е.м.}
  3. E_{\text{св}} = \Delta m\cdot 931E_{\text{св}} = 0{,}03038\cdot 931E_{\text{св}} = 28{,}28\,\text{МэВ}
  4. E'_{\text{св}} = \dfrac{E_{\text{св}}}{Z + N}E'_{\text{св}} = \dfrac{28{,}28}{4}E'_{\text{св}} = 7{,}07\,\tfrac{\text{МэВ}}{\text{нуклон}}

    Binding energy per nucleon is the average energy per nucleon needed to separate the nucleus completely, not the energy required to remove each individual nucleon.

Answer\Delta m = 0{,}03038\,\text{а.е.м.}E_{\text{св}} = 28{,}28\,\text{МэВ}E'_{\text{св}} = 7{,}07\,\tfrac{\text{МэВ}}{\text{нуклон}}

Position in this set: 8 from 20

Problem #84

Nuclear physics

How energy changes in reactions ^{4}_{2}He+^{14}_{7}N\to{}^{1}_{1}H+^{17}_{8}O and ^{4}_{2}He+^{9}_{4}Be\to{}^{1}_{0}n+^{12}_{6}C? Use atomic masses (amu): He-4 - 4.00260; N-14 - 14.00307; H-1 - 1.00783; O-17 - 16.99913; Be-9 - 9.01219; C-12 - 12. Neutron mass 1.00866 amu; 1 amu·c² = 931 MeV.

Answer options

  1. \left(0\,\text{МэВ};\ 0\,\text{МэВ}\right)
  2. \left(1{,}2\,\text{МэВ поглощается};\ 5{,}7\,\text{МэВ выделяется}\right)
  3. \left(1{,}2\,\text{МэВ выделяется};\ 5{,}7\,\text{МэВ поглощается}\right)
  4. \left(6{,}9\,\text{МэВ выделяется в обеих}\right)
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Given
\text{reaction 1}
^{4}_{2}\mathrm{He} + {}^{14}_{7}\mathrm{N} \to {}^{1}_{1}\mathrm{H} + {}^{17}_{8}\mathrm{O}
\text{reaction 2}
^{4}_{2}\mathrm{He} + {}^{9}_{4}\mathrm{Be} \to {}^{1}_{0}n + {}^{12}_{6}\mathrm{C}
Find

\Delta E_1\Delta E_2

  1. 4{,}00260 + 14{,}00307 = 18{,}00567;\quad 1{,}00783 + 16{,}99913 = 18{,}00696

    We compare the total masses before and after the reaction.

  2. \Delta m_1 = 18{,}00696 - 18{,}00567 = 0{,}00129\Delta m_1 = 0{,}00129\cdot 931\Delta E_1 = 1{,}2\,\text{МэВ}

    The mass has grown, which means energy has been absorbed.

  3. 4{,}00260 + 9{,}01219 = 13{,}01479;\quad 1{,}00866 + 12{,}00000 = 13{,}00866
  4. \Delta m_2 = 13{,}01479 - 13{,}00866 = 0{,}00613\Delta m_2 = 0{,}00613\cdot 931\Delta E_2 = 5{,}7\,\text{МэВ}

    The mass decreased - energy was released.

Answer\Delta E_1 = 1{,}2\,\text{МэВ поглощается}\Delta E_2 = 5{,}7\,\text{МэВ выделяется}

Position in this set: 9 from 20

Problem #85

Nuclear physics

How much energy will be released during complete fission 1\,\text{кг} uranium-235? How much uranium does a reactor consume in thermal power per year? 3200\,\text{МВт}? Accept the energy of one division 200\,\text{МэВ}.

Answer options

  1. \left(1{,}64 \cdot 10^{14}\,\text{Дж};\ 2470\,\text{кг}\right)
  2. \left(8{,}2 \cdot 10^{14}\,\text{Дж};\ 12350\,\text{кг}\right)
  3. \left(8{,}2 \cdot 10^{13}\,\text{Дж};\ 1235\,\text{кг}\right)
  4. \left(4{,}1 \cdot 10^{13}\,\text{Дж};\ 617{,}5\,\text{кг}\right)
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Given
m
1000\,\text{г}
M
235\,\tfrac{\text{г}}{\text{моль}}
E_1
200\,\text{МэВ}3{,}2\cdot 10^{-11}\,\text{Дж}
N_A
6\cdot 10^{23}\,\tfrac{1}{\text{моль}}
P
3200\,\text{МВт}3200\cdot 10^{6}\,\text{Вт}
t
1\,\text{год}365\cdot 24\cdot 3600\,\text{с}
Find

Em_2

  1. E = E_1 N = \dfrac{E_1 m N_A}{M}E = \dfrac{3{,}2\cdot 10^{-11}\cdot 1000\cdot 6\cdot 10^{23}}{235}E = 8{,}2\cdot 10^{13}\,\text{Дж}
  2. m_2 = m\dfrac{Pt}{E}m_2 = 1\,\text{кг}\cdot\dfrac{3200\cdot 10^{6}\cdot 365\cdot 24\cdot 3600}{\left(3{,}2\cdot 10^{-11}\cdot 1000\cdot 6\cdot 10^{23}\right)/235}m_2 = 1235\,\text{кг}

    A little more than a ton of uranium per year - against millions of tons of coal at the same power.

AnswerE = 8{,}2\cdot 10^{13}\,\text{Дж}m_2 = 1235\,\text{кг}

Position in this set: 10 from 20

Problem #86

Nuclear physics

Determine the energy yield of the reaction of the deuterium compound with tritium ^{2}_{1}H+^{3}_{1}H\to{}^{4}_{2}He+^{1}_{0}n. Use atomic masses (amu): H-2 - 2.01410; H-3 - 3.01605; He-4 - 4.00260. Neutron mass 1.00866 amu; 1 amu·c² = 931 MeV.

Answer options

  1. 8{,}8\,\text{МэВ}
  2. 35{,}2\,\text{МэВ}
  3. 176\,\text{МэВ}
  4. 17{,}6\,\text{МэВ}
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Given
\text{reaction}
^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + {}^{1}_{0}n
Find

\Delta E

  1. 2{,}01410 + 3{,}01605\Delta E = 5{,}03015\,\text{а.е.м.}

    Sum of masses before reaction.

  2. 4{,}00260 + 1{,}00866\Delta E = 5{,}01126\,\text{а.е.м.}

    Sum of masses after reaction.

  3. \Delta m = 5{,}03015 - 5{,}01126 = 0{,}01889\Delta m = 0{,}01889\cdot 931\Delta E = 17{,}6\,\text{МэВ}

    Thermonuclear fusion: mass lost, energy released.

Answer\Delta E = 17{,}6\,\text{МэВ}

Position in this set: 11 from 20

Problem #87

Nuclear physics

Determine the mass defect, total and specific binding energy of the oxygen-17 nucleus. The mass of the oxygen-17 atom is 16.99913 amu. For calculation, take the masses of the proton, neutron and electron equal to 1.00728; 1.00866 and 0.00055 amu; 1 amu·c² = 931 MeV.

Answer options

  1. \left(0{,}14145\,\text{а.е.м.};\ 131{,}7\,\text{МэВ};\ 7{,}75\,\text{МэВ/нуклон}\right)
  2. \left(0{,}070725\,\text{а.е.м.};\ 65{,}85\,\text{МэВ};\ 3{,}875\,\text{МэВ/нуклон}\right)
  3. \left(0{,}2829\,\text{а.е.м.};\ 263{,}4\,\text{МэВ};\ 15{,}5\,\text{МэВ/нуклон}\right)
  4. \left(1{,}4145\,\text{а.е.м.};\ 1317\,\text{МэВ};\ 77{,}5\,\text{МэВ/нуклон}\right)
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Given
\text{core}
^{17}_{8}\mathrm{O}
m_a
16{,}99913\,\text{а.е.м.}
Z
8
N
9
m_p
1{,}00728\,\text{а.е.м.}
m_n
1{,}00866\,\text{а.е.м.}
m_e
0{,}00055\,\text{а.е.м.}
Find

\Delta mE_{\text{св}}E'_{\text{св}}

  1. m_{\text{я}} = m_a - Zm_em_{\text{я}} = 16{,}99913 - 8\cdot 0{,}00055m_{\text{я}} = 16{,}99473\,\text{а.е.м.}
  2. \Delta m = Zm_p + Nm_n - m_{\text{я}}\Delta m = 8\cdot 1{,}00728 + 9\cdot 1{,}00866 - 16{,}99473\Delta m = 0{,}14145\,\text{а.е.м.}
  3. E_{\text{св}} = \Delta m\cdot 931E_{\text{св}} = 0{,}14145\cdot 931E_{\text{св}} = 131{,}7\,\text{МэВ}
  4. E'_{\text{св}} = \dfrac{E_{\text{св}}}{Z + N}E'_{\text{св}} = \dfrac{131{,}7}{17}E'_{\text{св}} = 7{,}75\,\tfrac{\text{МэВ}}{\text{нуклон}}

Answer\Delta m = 0{,}14145\,\text{а.е.м.}E_{\text{св}} = 131{,}7\,\text{МэВ}E'_{\text{св}} = 7{,}75\,\tfrac{\text{МэВ}}{\text{нуклон}}

Position in this set: 12 from 20

Problem #88

Nuclear physics

Available 40\,\text{мг} drug with half-life 8 days. How long will it take to remain 5\,\text{мг}? What fraction of atoms decays in a time equal to half the half-life?

Answer options

  1. \left(240\,\text{суток};\ 2{,}93\,\text{доля}\right)
  2. \left(24\,\text{суток};\ 0{,}293\,\text{доля}\right)
  3. \left(12\,\text{суток};\ 0{,}1465\,\text{доля}\right)
  4. \left(48\,\text{суток};\ 0{,}586\,\text{доля}\right)
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Given
m_0
40\,\text{мг}
T
8\,\text{сут}
m
5\,\text{мг}
t_2
\dfrac{T}{2}
Find

t_1\dfrac{m_0 - m}{m_0}

  1. m = m_0\cdot 2^{-t/T}\ \Rightarrow\ 2^{t/T} = \dfrac{m_0}{m}\dfrac{40}{5} = 8 = 2^{3}8=2^3

    The remaining mass is 8 times less than the initial one: these are three half-lives, t/T = 3.

  2. t_1 = 3Tt_1 = 3\cdot 8t_1 = 24\,\text{сут}
  3. \dfrac{m_0 - m}{m_0} = 1 - 2^{-t_2/T} = 1 - 2^{-1/2}\dfrac{m_0 - m}{m_0} = 1 - \dfrac{1}{\sqrt{2}}\dfrac{m_0 - m}{m_0} = 0{,}293 = 29{,}3\%

    Over half the half-life, approximately 29 percent decays.

Answert_1 = 24\,\text{сут}\dfrac{m_0 - m}{m_0} = 0{,}293

Position in this set: 13 from 20

Problem #89

Nuclear physics

Determine the energy output of a nuclear reaction ^{2}_{1}H+^{7}_{3}Li\to{}^{1}_{0}n+2{}^{4}_{2}He. Use atomic masses (amu): H-2 - 2.01410; Li-7 - 7.01601; He-4 - 4.00260. Neutron mass 1.00866 amu; 1 amu·c² = 931 MeV.

Answer options

  1. 30{,}26\,\text{МэВ}
  2. 151{,}3\,\text{МэВ}
  3. 15{,}13\,\text{МэВ}
  4. 7{,}565\,\text{МэВ}
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Given
\text{reaction}
^{2}_{1}\mathrm{H}+{}^{7}_{3}\mathrm{Li}\to{}^{1}_{0}n+2{}^{4}_{2}\mathrm{He}
Find

\Delta E

  1. 2{,}01410 + 7{,}01601\Delta E = 9{,}03011\,\text{а.е.м.}
  2. 1{,}00866+2\cdot4{,}00260\Delta E = 9{,}01386\,\text{а.е.м.}
  3. \Delta m=9{,}03011-9{,}01386=0{,}01625\Delta m = 0{,}01625\cdot931\Delta E = 15{,}13\,\text{МэВ}

Answer\Delta E=15{,}13\,\text{МэВ}

Position in this set: 14 from 20

Problem #91

Nuclear physics

Determine the mass defect and binding energy of the boron-11 nucleus. The mass of a boron-11 atom is 11.00931 amu. For calculation, take the masses of the proton, neutron and electron equal to 1.00728; 1.00866 and 0.00055 amu; 1 amu·c² = 931 MeV.

Answer options

  1. \left(0{,}0818\,\text{а.е.м.};\ 76{,}2\,\text{МэВ}\right)
  2. \left(0{,}0409\,\text{а.е.м.};\ 38{,}1\,\text{МэВ}\right)
  3. \left(0{,}1636\,\text{а.е.м.};\ 152{,}4\,\text{МэВ}\right)
  4. \left(0{,}818\,\text{а.е.м.};\ 762\,\text{МэВ}\right)
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Given
\text{core}
^{11}_{5}\mathrm{B}
m_a
11{,}00931\,\text{а.е.м.}
Z
5
N
6
m_p
1{,}00728\,\text{а.е.м.}
m_n
1{,}00866\,\text{а.е.м.}
m_e
0{,}00055\,\text{а.е.м.}
Find

\Delta mE_{\text{св}}

  1. m_{\text{я}} = m_a - Zm_em_{\text{я}} = 11{,}00931 - 5\cdot 0{,}00055m_{\text{я}} = 11{,}00656\,\text{а.е.м.}
  2. \Delta m = Zm_p + Nm_n - m_{\text{я}}\Delta m = 5\cdot 1{,}00728 + 6\cdot 1{,}00866 - 11{,}00656\Delta m = 0{,}0818\,\text{а.е.м.}
  3. E_{\text{св}} = \Delta m\cdot 931E_{\text{св}} = 0{,}0818\cdot 931E_{\text{св}} = 76{,}2\,\text{МэВ}

Answer\Delta m = 0{,}0818\,\text{а.е.м.}E_{\text{св}} = 76{,}2\,\text{МэВ}

Position in this set: 15 from 20

Problem #146

Atomic nucleus and radioactivity

Nucleus mass number 23, charge number 11. How many neutrons are in the nucleus?

Answer options

  1. 12\,\text{нейтронов}
  2. 24\,\text{нейтронов}
  3. 6\,\text{нейтронов}
  4. 36\,\text{нейтронов}
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Given
A
23\,\text{нуклонов}
Z
11\,\text{протонов}
Find

N

  1. N=A-ZN=23 - 11N=12\,\text{нейтронов}

    The mass number equals the total number of protons and neutrons.

Answer12\,\text{нейтронов}

Position in this set: 16 from 20

Problem #147

Atomic nucleus and radioactivity

Nucleus mass number 27, charge number 13. How many neutrons are in the nucleus?

Answer options

  1. 28\,\text{нейтронов}
  2. 14\,\text{нейтронов}
  3. 7\,\text{нейтронов}
  4. 42\,\text{нейтронов}
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Given
A
27\,\text{нуклонов}
Z
13\,\text{протонов}
Find

N

  1. N=A-ZN=27 - 13N=14\,\text{нейтронов}

    The mass number equals the total number of protons and neutrons.

Answer14\,\text{нейтронов}

Position in this set: 17 from 20

Problem #148

Atomic nucleus and radioactivity

Nucleus with mass number 238 and charge number 92 released one alpha particle. How many neutrons are in the daughter nucleus?

Answer options

  1. 288\,\text{нейтронов}
  2. 144\,\text{нейтронов}
  3. 72\,\text{нейтронов}
  4. 432\,\text{нейтронов}
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Given
A
238\,\text{нуклонов}
Z
92\,\text{протонов}
Find

N

  1. A_2=A-4A_2=238 - 4A_2=234\,\text{нуклонов}

    An alpha particle carries away four nucleons.

  2. Z_2=Z-2Z_2=92 - 2Z_2=90\,\text{протонов}

    Of these, two are protons.

  3. N=A_2-Z_2N=234 - 90N=144\,\text{нейтронов}

    The number of neutrons also decreases by two.

Answer144\,\text{нейтронов}

Position in this set: 18 from 20

Problem #149

Atomic nucleus and radioactivity

Nucleus with mass number 226 and charge number 88 released one alpha particle. How many neutrons are in the daughter nucleus?

Answer options

  1. 272\,\text{нейтронов}
  2. 68\,\text{нейтронов}
  3. 136\,\text{нейтронов}
  4. 408\,\text{нейтронов}
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Given
A
226\,\text{нуклонов}
Z
88\,\text{протонов}
Find

N

  1. A_2=A-4A_2=226 - 4A_2=222\,\text{нуклонов}

    An alpha particle carries away four nucleons.

  2. Z_2=Z-2Z_2=88 - 2Z_2=86\,\text{протонов}

    Of these, two are protons.

  3. N=A_2-Z_2N=222 - 86N=136\,\text{нейтронов}

    The number of neutrons also decreases by two.

Answer136\,\text{нейтронов}

Position in this set: 19 from 20

Problem #150

Atomic nucleus and radioactivity

The sample contained 8000 radioactive nuclei. After three half-lives, how many nuclei have decayed? The sample is not replenished.

Answer options

  1. 14000\,\text{ядер}
  2. 3500\,\text{ядер}
  3. 7000\,\text{ядер}
  4. 21000\,\text{ядер}
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Given
N
8000\,\text{ядер}
Find

D

  1. N_1=\frac{N}{2}N_1=\frac{8000}{2}N_1=4000\,\text{ядер}

    After the first period, half of the nuclei remain.

  2. N_3=\frac{N_1}{2^2}N_3=\frac{4000}{4}N_3=1000\,\text{ядер}

    Over two more periods, the amount decreases fourfold.

  3. D=N-N_3D=8000 - 1000D=7000\,\text{ядер}

    The number of decayed nuclei is required.

Answer7000\,\text{ядер}

Position in this set: 20 from 20

Problem #151

Atomic nucleus and radioactivity

The sample contained 12000 radioactive nuclei. After three half-lives, how many nuclei have decayed? The sample is not replenished.

Answer options

  1. 21000\,\text{ядер}
  2. 5250\,\text{ядер}
  3. 31500\,\text{ядер}
  4. 10500\,\text{ядер}
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Given
N
12000\,\text{ядер}
Find

D

  1. N_1=\frac{N}{2}N_1=\frac{12000}{2}N_1=6000\,\text{ядер}

    After the first period, half of the nuclei remain.

  2. N_3=\frac{N_1}{2^2}N_3=\frac{6000}{4}N_3=1500\,\text{ядер}

    Over two more periods, the amount decreases fourfold.

  3. D=N-N_3D=12000 - 1500D=10500\,\text{ядер}

    The number of decayed nuclei is required.

Answer10500\,\text{ядер}

The theory behind these problems

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Atomic nucleus and radioactivityNuclear composition, mass defect, binding energy and the law of radioactive decay.

Formulas by topic

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PeriodT · сEnergyE, W · Дж