Question:
The de-Broglie wavelength of an electron moving in the \(n^{th}\) Bohr orbit of radius 'r' is
The de-Broglie wavelength of an electron moving in the \(n^{th}\) Bohr orbit of radius 'r' is
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In the nth Bohr orbit the circumference holds n de Broglie wavelengths.
Updated On: Oct 1, 2026
- \(nπr\)
- \(\frac{nr}{π}\)
- \(\frac{2πr}{n}\)
- \(\frac{nr}{2π}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understand the concept
De Broglie explained Bohr's quantisation by saying that the electron's standing wave must fit the orbit: the circumference of the orbit equals a whole number of wavelengths.
Step 2: Write the condition
\[ 2\pi r = n\lambda \]
Step 3: Solve
\[ \lambda = \frac{2\pi r}{n} \]
Step 4: Result
Option (C). This is equivalent to Bohr's angular momentum condition \(mvr = \dfrac{nh}{2\pi}\) with \(\lambda = \dfrac{h}{mv}\).
Final Answer:
The wavelength is 2 pi r / n. This is option (C). \[ \boxed{\text{(C) }\frac{2\pi r}{n}} \]
De Broglie explained Bohr's quantisation by saying that the electron's standing wave must fit the orbit: the circumference of the orbit equals a whole number of wavelengths.
Step 2: Write the condition
\[ 2\pi r = n\lambda \]
Step 3: Solve
\[ \lambda = \frac{2\pi r}{n} \]
Step 4: Result
Option (C). This is equivalent to Bohr's angular momentum condition \(mvr = \dfrac{nh}{2\pi}\) with \(\lambda = \dfrac{h}{mv}\).
Final Answer:
The wavelength is 2 pi r / n. This is option (C). \[ \boxed{\text{(C) }\frac{2\pi r}{n}} \]
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