Question

The Bohr orbit radius for the hydrogen atom (n = 1) is
approximately 0.530 Å. The radius for the first excited state
(n = 2) orbit is (in Å)
(a) 0.13 (b) 1.06 (c) 4.77 (d) 2.12

Answer 1

We have radius as:
$\frac{ {n}^{2} }{z} \times0. 530 \: Å$

Where n is the Principal quantum number of orbital and z is the atomic number.

z= 1
When n = 1 we have,

orbit radius = 0.530 Å.


when n = 2 (and z still = 1)

orbit radius=
$\frac{ {2}^{2} }{1} \times 0.530 \: Å\\ \\ = 4 \times 0.530 \: Å \\ \\ = 2.120 \: Å$


Therefore option D is your answer.

Answer 2

The radius for the first excited state of hydrogen atom is 2.12 Å

Explanation:

From the formula of Bohr's radius, $r=\frac{0.53 n^{2}}{Z}$

Here n is the energy level and Z is the atomic number

For the given state n =2

And Z = 1 for Hydrogen

Therefore, $r=\frac{0.53 n^{2}}{Z}=\frac{0.53\times 4}{1} =2.12$

The radius of first excited state is 2.12

Therefore the correct answer is option (d)