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7. (a) Which of these describes isotopes of an element? A same atomic number different number of neutrons B same atomic number different number of protons C same mass number different number of neutrons D same mass number different number of protons (b) Figure 9 represents a decay that can happen inside the nucleus of an atom - Edexcel - GCSE Physics - Question 7 - 2020 - Paper 1

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7.-(a)-Which-of-these-describes-isotopes-of-an-element?--A---same-atomic-number---different-number-of-neutrons-B---same-atomic-number---different-number-of-protons-C---same-mass-number---different-number-of-neutrons-D---same-mass-number---different-number-of-protons--(b)-Figure-9-represents-a-decay-that-can-happen-inside-the-nucleus-of-an-atom-Edexcel-GCSE Physics-Question 7-2020-Paper 1.png

7. (a) Which of these describes isotopes of an element? A same atomic number different number of neutrons B same atomic number different number of protons C... show full transcript

Worked Solution & Example Answer:7. (a) Which of these describes isotopes of an element? A same atomic number different number of neutrons B same atomic number different number of protons C same mass number different number of neutrons D same mass number different number of protons (b) Figure 9 represents a decay that can happen inside the nucleus of an atom - Edexcel - GCSE Physics - Question 7 - 2020 - Paper 1

Step 1

Which of these describes isotopes of an element?

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Answer

The correct answer is A: same atomic number, different number of neutrons. This definition accurately reflects the properties of isotopes, which are variants of elements that differ in the number of neutrons while maintaining the same number of protons.

Step 2

Which decay is represented in Figure 9?

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Answer

The decay represented in Figure 9 is C: beta plus. In this decay process, a proton is transformed into a neutron and a positron, which is consistent with the illustration given.

Step 3

Estimate the activity of this source in the year 2020.

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Answer

To estimate the activity in the year 2020, we recognize that 20 years have passed since 2000. Since the half-life of cobalt-60 is 5 years, this corresponds to 4 half-lives (20/5 = 4).

Using the formula for activity after n half-lives:

A=A0(12)nA = A_0 \left(\frac{1}{2}\right)^n

where:

  • A0=38.5A_0 = 38.5 kBq (initial activity)
  • n=4n = 4 (number of half-lives)

Thus:

A=38.5(12)4=38.5×116=2.40625 kBqA = 38.5 \left(\frac{1}{2}\right)^4 = 38.5 \times \frac{1}{16} = 2.40625 \text{ kBq}

Rounded to three significant figures, the activity in the year 2020 is approximately 2.4 kBq.

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