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nekit [7.7K]
3 years ago
8

Part of one row of the periodic table is shown. Which element’s atoms have the greatest average number of neutrons?

Physics
2 answers:
Zanzabum3 years ago
8 0

Number of neutrons of the elements can be calculated if the atomic number and mass number is known.

As, Number of neutrons= Mass number- Atomic number

For Tin(sn), the mass number is 119 and atomic number is 50, then

Number of neutrons= 119-50=69

For Antimony(sb), the mass number is 122 and the atomic number is 51, then

Number of neutrons=122-51=71

For Tellurium(te), the mass number is 128 and atomic number is 52,then

Number of neutrons=128-52=76

Fro Iodine(I), the mass number is 127 and the atomic number is 53, then

Number of neutrons=127-53=74

Since, the number of neutrons are large in Tellurium, therefore Tellurium have the greatest average number of neutrons.

tekilochka [14]3 years ago
6 0

The answer is C. This can be derived from the mass number of the elements (which is the number indicated below the elements; the number above the elements is the atomic number--number of protons). The mass number is the total number of protons and neutrons in the nucleus of the atom. The higher the protons and neutrons numbers, the higher the mass number. Most often, neutrons and protons occur in same numbers in the nucleus, but occasionally, one can exceed the other by one or two in isotopes.

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Explanation

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation to find the force

\begin{gathered} F=qvBsinθ \\ where\text{ F is the magnetic force} \\ q\text{ is the charge} \\ v\text{ is the velocity of the charge} \\ Bis\text{ the magnetic field} \\ \theta\text{ is the angle} \end{gathered}

so

Step 1

Let

\begin{gathered} q=1.25*10^{-4\text{ }}C \\ v=5200\frac{m}{s} \\ \theta=37\text{ \degree} \\ B=8.49*10^{-4}T \end{gathered}

now, replace

\begin{gathered} F=qvBs\imaginaryI n\theta \\ F=1.25*10^{-4}\text{ C*5200 }\frac{m}{s}*8.49*10^{-4}Tsin(37) \\ F=0.0003321\text{ Newtons} \\  \end{gathered}

so, the answer is

0.0003321 Newtons

I hope this helps you

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