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Write the symbol (in the form \(_{Z}^{A} \mathrm{X}\) ) for the nuclide that has 78 neutrons and 53 protons.

Short Answer

Expert verified
Answer: The nuclide symbol for this isotope is \(_{53}^{131} \mathrm{I}\).

Step by step solution

01

Find the element symbol

The element symbol can be found by matching the atomic number (Z) to its corresponding element in the periodic table. The problem states that there are 53 protons, so the atomic number is 53. Upon looking it up on the periodic table, the element with Z = 53 is Iodine (I).
02

Calculate the mass number

To find the mass number (A), add the number of protons (Z) and the number of neutrons. In this case, there are 53 protons and 78 neutrons. The mass number is then: A = 53 (protons) + 78 (neutrons) = 131
03

Write the nuclide symbol

Using the values found in steps 1 and 2, the nuclide symbol in the form \(_{Z}^{A} \mathrm{X}\) can be written as: \(_{53}^{131} \mathrm{I}\)

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Most popular questions from this chapter

The half-life of I-131 is 8.0 days. A sample containing I- 131 has an activity of \(6.4 \times 10^{8}\) Bq. How many days later will the sample have an activity of \(2.5 \times 10^{6} \mathrm{Bq} ?\) (Wills tutorial: radioactive decay)
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