Question

the radius of hydrogen atom in the ground state is 52.9 pm. the radius of the first excited state of he+ ion will be​

Answer 1

this is the correct answer

Answer 2

Given:

The radius of hydrogen atoms in the ground state is 52.9 pm.

To Find:

The radius of the first excited state of he+ ion.

Solution:

To find the we will follow the following steps:

As we know,

He+ is a hydrogen an atom and the Radius of hydrogen species is given by the formula:

$radius = 4\pi \: eo \frac{ {n}^{2} {z} }{mz^{2} {e}^{2} }$

eo= permittivity of free space

n = energy level

m = electron rest mass

e = elementary charge

The value of tex]4\pi \: eo \frac{1}{m{ {e}^{2} } [/tex] = 52.9pm.

So, the radius is directly proportional to

$52.9 \times \frac{ {n}^{2} }{z} pm$

The radius of He+ ion =

$52.9 \times \frac{ {2}^{2} }{2} = 52.9\times 2 = 105.8 \: pm$

Henceforth, The radius of the first excited state of the he+ ion is 105.8pm.