The de-Broglie's wavelength of an electron in the \( 4^{\text{th}} \) orbit is ______ \( \pi a_0 \). (\( a_0 = \text{Bohr's radius} \))
Correct Answer: 8
Approach Solution - 1
The condition for the de-Broglie wavelength of an electron in an orbit is given by:
\[ 2 \pi r_n = n \lambda_d \]
where \( r_n \) is the radius of the \( n \)-th orbit, \( n \) is the principal quantum number, and \( \lambda_d \) is the de-Broglie wavelength.
The radius of the \( n \)-th orbit in the Bohr model is given by:
\[ r_n = 2 \pi a_0 \frac{n^2}{Z} \]
Substituting this into the equation:
\[ 2 \pi a_0 \frac{n^2}{Z} = n \lambda_d \]
For an electron in the 4th orbit and for a hydrogen atom (\( Z = 1 \)):
\[ 2 \pi a_0 \frac{4^2}{1} = 4 \lambda_d \]
Simplifying:
\[ \lambda_d = 8 \pi a_0 \]
Hence, the de-Broglie's wavelength of the electron in the 4th orbit is \( 8 \pi a_0 \).
Approach Solution -2
According to Bohr’s theory, the angular momentum of an electron in the \( n^{\text{th}} \) orbit is quantized and given by:
\[ mvr = \frac{nh}{2\pi} \] The de-Broglie wavelength of the electron is given by: \[ \lambda = \frac{h}{mv} \]
Step 2: Substitute the value of \( mv \) from Bohr’s equation.
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n} \]
Step 3: Relation between radius and principal quantum number.
For a hydrogen atom: \[ r_n = n^2 a_0 \] Substitute this value into the wavelength formula: \[ \lambda = \frac{2\pi (n^2 a_0)}{n} = 2\pi n a_0 \]
Step 4: For the 4th orbit.
\[ \lambda_4 = 2\pi (4) a_0 = 8\pi a_0 \]
Step 5: Final Answer.
\[ \boxed{8} \] Hence, the de-Broglie wavelength of an electron in the 4th orbit is \( 8\pi a_0 \).
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