The radius of fifth orbit of the $\mathrm{Li}^{++}$ is $\_\_\_\_$ $\times 10^{-12}$ m. Take : radius of hydrogen atom $=0.51 \AA$

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Modern Physics

The radius of fifth orbit of the

Li^{++}

____

\times 1 0^{- 12}

m. Take : radius of hydrogen atom

= 0.51 \overset{˚}{A}

Answer: 425

Solution

For hydrogen-like ions, the radius of the

n

-th orbit is given by

r_{n} = r_{0} \frac{n ^{2}}{Z}

, where

r_{0} = 0.51 \overset{˚}{A}

is the Bohr radius,

n

is the principal quantum number, and

Z

is the atomic number. For

Li^{++}

Z = 3

and for the fifth orbit,

n = 5

. Thus

r_{5} = 0.51 \times \frac{25}{3} = 4.25 \overset{˚}{A} = 4.25 \times 1 0^{- 10}

= 425 \times 1 0^{- 12}

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Integer answer

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