Chemistry, asked by muralireddy97, 11 months ago

the ratio between total acceleration of the electron in singly ionized helium atom and doubly ionized lithium atom is​

Answers

Answered by panesarh989
0

Answer:

ANSWER

According Bohr's model of an atom V_n=2.18\times 10^\dfrac{z}{n}\propto \dfrac{z}{n}V

n

=2.18×10

n

z

n

z

r_n=0.529\dfrac{n^2}{z}\propto \dfrac{n^2}{z}r

n

=0.529

z

n

2

z

n

2

The angular acceleration of the electron a=\dfrac{v^2_n}{r_n}\propto\dfrac{z^3}{n^4}a=

r

n

v

n

2

n

4

z

3

for ground state n=1n=1

\Rightarrow a\propto z^3⇒a∝z

3

\dfrac{a_{He}}{a_{Li}}=\dfrac{z^3_{He}}{z^3_{Li}}=\dfrac{z^3}{3^3}=\dfrac{8}{27}

a

Li

a

He

=

z

Li

3

z

He

3

=

3

3

z

3

=

27

8

z_{He}=2z

He

=2

z_{Li}=3z

Li

=3.

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