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GRADE 9 PRACTICE

Physics Problems: Atomic structure, Grade 9

Atomic models, spectra, radioactivity and nuclear composition.

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

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Problem statements

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9 problems
Position in this set: 1 from 9

Problem #55

Atom, nucleus, spectra and radioactivity

What concept corresponds to the description: “The nucleus consists of protons and neutrons, collectively called nucleons and held together by nuclear forces.”?

Answer options

  1. \text{Composition of the atomic nucleus}
  2. \text{Atomic models and spectra}
  3. \text{Radioactivity and detection}
  4. \text{Point particle}
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Given
\text{Description}
\text{The nucleus is made up of protons and neutrons, collectively called nucleons, held together by nuclear forces.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Z counts protons and identifies the element; A counts protons plus neutrons. Isotopes differ in neutron number. A hydrogen-1 nucleus contains just one proton; not every nucleus must contain neutrons.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Composition of the atomic nucleus}

Answer\text{Concept} = \text{Composition of the atomic nucleus}

Position in this set: 2 from 9

Problem #56

Atom, nucleus, spectra and radioactivity

What concept does the description correspond to: “The nuclear model explains the scattering of alpha particles, and the discrete energy states of the atom explain the line emission and absorption spectra.”?

Answer options

  1. \text{Atomic models and spectra}
  2. \text{Composition of the atomic nucleus}
  3. \text{Radioactivity and detection}
  4. \text{Point particle}
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Given
\text{Description}
\text{The nuclear model explains alpha-particle scattering; discrete atomic energy states explain emission and absorption line spectra.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    An atom emits or absorbs a photon when moving between allowed energy states.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Atomic models and spectra}

Answer\text{Concept} = \text{Atomic models and spectra}

Position in this set: 3 from 9

Problem #57

Atomic structure

A neutral atom has a charge number of 13 and a mass number of 27. How many protons, electrons and neutrons does it have?

Answer options

  1. 13\,\text{p};\ 13\,\text{e};\ 14\,\text{n}
  2. 13\,\text{p};\ 14\,\text{e};\ 27\,\text{n}
  3. 27\,\text{p};\ 13\,\text{e};\ 14\,\text{n}
  4. 14\,\text{p};\ 13\,\text{e};\ 13\,\text{n}
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Given
Z
13
A
27
q_{\text{атома}}
0
Find

N_pN_eN_n

  1. N_p=ZN_p = 13
  2. N_e=N_pN_e = 13

    The atom is neutral.

  3. N_n=A-ZN_n=27-13N_n = 14

Answer13\,\text{протонов}13\,\text{электронов}14\,\text{нейтронов}

Position in this set: 4 from 9

Problem #172

Atom, nucleus, spectra 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: 5 from 9

Problem #173

Atom, nucleus, spectra 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: 6 from 9

Problem #174

Atom, nucleus, spectra 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: 7 from 9

Problem #175

Atom, nucleus, spectra 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: 8 from 9

Problem #176

Atom, nucleus, spectra 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: 9 from 9

Problem #177

Atom, nucleus, spectra 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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Atom, nucleus, spectra and radioactivityAtomic models, nucleons, isotopes, radiation and particle detection.

Formulas by topic

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PeriodT · с