In Bohr's model of the hydrogen atom, the radius of a stationary orbit is proportional to … — JAMB UTME

In Bohr's model of the hydrogen atom, the radius of a stationary orbit is proportional to $n^2$ , where $n$ is the principal quantum number. If the radius of the ground-state orbit ( $n = 1$ ) is $r_1$ , what is the radius of the orbit corresponding to the second excited state?

A
$3r_1$
B
$4r_1$
C
$6r_1$
$9r_1$ Answer

Answer

The radius of the electron's orbit in the second excited state is

9r_1

In Bohr's atomic model, the ground state corresponds to the quantum number

n = 1

. The excited states are numbered sequentially above the ground state: the first excited state is

n = 2

and the second excited state is

n = 3

. Since the orbital radius scales with the square of the principal quantum number (

r_n = n^2 r_1

), substituting

n = 3

yields

r_3 = 3^2 r_1 = 9r_1

. Thus, the option stating

9r_1

is correct.

Step-by-Step Solution

Identify the principal quantum number

n

for the second excited state.

n = 3

The ground state corresponds to

n = 1

, the first excited state corresponds to

n = 2

, and the second excited state corresponds to

n = 3

Apply Bohr's radius formula for hydrogen-like atoms.

r_n = n^2 r_1

According to Bohr's postulates, the orbital radius is proportional to the square of the principal quantum number

n

Calculate

r_3

by substituting

n = 3

into the radius expression.

r_3 = 3^2 r_1 = 9r_1

Squaring

n = 3

gives

9

, making the radius 9 times the ground-state radius.

Key Concept

Bohr's quantization of orbital radius (

r_n \propto n^2

)

Estimated Time:1m 0s

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