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grigory [225]
3 years ago
7

How do the valence electron configurations of alkali metals compared with each other?

Chemistry
1 answer:
melisa1 [442]3 years ago
8 0
They have the same number of valence electrons
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Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10
kkurt [141]

Answer:

The diameter of the hydrogen \mathbf{d =1.0605 \times 10^{-10}\ m}

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

L = \dfrac{nh}{2 \pi}

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

mvr= \dfrac{nh}{2 \pi}

v= \dfrac{nh}{2 \pi mr} ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}   ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2

ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}

r =\dfrac{n^2h^2}{4 \pi^2 mke^2}

For ground-state n = 1

h = (6.625 \times 10^{-34} \ J.s)^2

m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)

Ke = (1.6 \times 10^{-19} \ C)^2

r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}

r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}

\mathbf{r = 5.3025 \times 10^{-11} \ m}

Therefore, the diameter of hydrogen d = 2r

\mathbf{d = ( 2 \times  5.3025 \times 10^{-11} \ m})}

\mathbf{d =1.0605 \times 10^{-10}\ m}}

4 0
3 years ago
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