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3 years ago

Does atomic radius increase or decrease as you go down a group/family on periodic table?

Chemistry

1 answer:

SVEN [57.7K]3 years ago

7 0

The atomic radius increases as we go down a group/family on a periodic table.

Explanation:

"Atomic radius" is defined as the "one-half of the distance" between the "nuclei of two atoms". While moving down the group in a periodic table there is an "increase in quantum number n" which allows the valence electrons to occupy on higher levels. This shows that the valence electrons move away from the nucleus, which makes them to be loosely held. Thus, atomic radius is large.

Similarly while moving across the periodic table from "left to right" the atomic radius gradually decreases.

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How many moles are in 0.423 g of Ca

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16.952994

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On occasion, it has been found that the oxidation of borneol doesn't go to completion (possibly because of poor stirring or insu

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Answer:

Check the explanation

Explanation:

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3 years ago

The work function for metallic magnesium is 3.68 eV. Calculate the velocity in km/s for electrons ejected from a metallic magnes

Lemur [1.5K]

Answer:

v = 301.62 km/s

Explanation:

Given that:

The work function of the magnesium = 3.68 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $3.68\times 1.6022\times 10^{-19}\ J=5.8961\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=315\ nm=315\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{315\times 10^{-9}}=5.8961\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$5.8961\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{315\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=6.31047619\times 10^{-19}-5.8961\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=0.41437619\times 10^{-19}$

$v^2=0.0909717212\times 10^{12}$

v = 3.0162 × 10⁵ m/s

Also, 1 m = 0.001 km

So, v = 301.62 km/s

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