Question:
The radius of first Bohr orbit of hydrogen atom is \(r_o\)\( \AA\). The wavelength (in \(\AA\)) of electron associated with sixth orbit of same atom is
The radius of first Bohr orbit of hydrogen atom is \(r_o\)\( \AA\). The wavelength (in \(\AA\)) of electron associated with sixth orbit of same atom is
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For hydrogen atom,
\[
r_n=n^2r_o
\]
and
\[
2\pi r_n=n\lambda_n
\]
Combining both,
\[
\lambda_n=2\pi nr_o
\]
Thus for \(n=6\),
\[
\lambda_6=12\pi r_o/2=6\pi r_o.
\]
Updated On: Jun 22, 2026
- \(6\pi r_o\)
- \(3\pi r_o\)
- \(8\pi r_o\)
- \(12\pi r_o\) \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
According to Bohr's quantization condition, an electron can revolve around the nucleus only in those circular orbits for which the circumference of the orbit contains an integral number of de Broglie wavelengths.
\[
2\pi r_n=n\lambda_n
\]
where
\[
r_n=n^2r_o
\]
and \(r_o\) is the radius of the first Bohr orbit.
This relation directly connects Bohr's atomic model with de Broglie's wave theory.
Step 1: Write the radius of the sixth Bohr orbit.
For hydrogen atom, \[ r_n=n^2r_o \] For \(n=6\), \[ r_6=6^2r_o \] \[ r_6=36r_o \] Thus the radius of the sixth orbit is \[ \boxed{r_6=36r_o} \]
Step 2: Apply Bohr's quantization condition.
The de Broglie wavelength associated with the electron in the \(n^{th}\) orbit is obtained from \[ 2\pi r_n=n\lambda_n \] Substituting \(n=6\), \[ 2\pi r_6=6\lambda_6 \] Substituting \(r_6=36r_o\), \[ 2\pi(36r_o)=6\lambda_6 \] \[ 72\pi r_o=6\lambda_6 \] \[ \lambda_6=12\pi r_o \] Dividing by 2, \[ \lambda_6=6\pi r_o \]
Step 3: Obtain the required wavelength.
Therefore the wavelength associated with the electron in the sixth orbit is \[ \boxed{\lambda_6=6\pi r_o} \] Hence the correct option is \[ \boxed{\text{(A)}} \]
Step 1: Write the radius of the sixth Bohr orbit.
For hydrogen atom, \[ r_n=n^2r_o \] For \(n=6\), \[ r_6=6^2r_o \] \[ r_6=36r_o \] Thus the radius of the sixth orbit is \[ \boxed{r_6=36r_o} \]
Step 2: Apply Bohr's quantization condition.
The de Broglie wavelength associated with the electron in the \(n^{th}\) orbit is obtained from \[ 2\pi r_n=n\lambda_n \] Substituting \(n=6\), \[ 2\pi r_6=6\lambda_6 \] Substituting \(r_6=36r_o\), \[ 2\pi(36r_o)=6\lambda_6 \] \[ 72\pi r_o=6\lambda_6 \] \[ \lambda_6=12\pi r_o \] Dividing by 2, \[ \lambda_6=6\pi r_o \]
Step 3: Obtain the required wavelength.
Therefore the wavelength associated with the electron in the sixth orbit is \[ \boxed{\lambda_6=6\pi r_o} \] Hence the correct option is \[ \boxed{\text{(A)}} \]
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