Question

The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of.?

Answer 1

Answer:

where , K = 1/4πε0 = constant.

mvr = nh/2π

v = nh/2πmr ——–eq 2.

KZe2 /r = m(nh/2πmr)2

h = 6.625 x 10-34 , π = 3.14 , m = 9.1 x 10-31, e = 1.6 x 10-19 , K = 9 x 109 Nm2/C2

r = n2 x 6.625 x 10-34 x 6.625 x 10-34 / 4 x 3.14 x 3.14 x 9.1 x 10-31 x 1.6 x 10-19 x 1.6 x 10-19 x 9 x 109

Answer 2

Concept introduction:

Each circling electron in classical physics is described by a wave function, a mathematical expression that may be compared to a pulsating guitar string spread out along the path of the electron's orbit. Orbitals are the name for these harmonics. See also Bohr's hypothesis of the atom in quantum mechanics.

Explanation:

We have to tell that The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of what.

A question regarding this has been given to us.

The radius of the $4$ th orbit of the electron will be smaller than its $8$th orbit by a factor of $4$.

$r4 = 0.529 *107 * 4^{2}$

$r_{s}= 0.529 * 10^{-8}* 8^{2}$

$\frac{r_{s}}{r_{4}} =4$

Final answer:

So, the final answer of the question is $4$.

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