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4 years ago

6

Where on the table would a group’s ionization energy be the greatest?

1 answer:

ANEK [815]4 years ago

6 0

Explanation:

The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest.

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Explanation:

It is given that,

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$\dfrac{1}{2}mv^2=qV$

$v=\sqrt{\dfrac{2qV}{m}}$

q is the charge on an electron

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}$

v = 78608.58 m/s

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To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

$qvB=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}$

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

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Answer:

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