Arrange the following cases of the electron moving in different orbits in hydrogen atom in increasing order of their de Broglie wavelengths.
A. n = 1
B. n = 3
C. n = 4
D. n = 2
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A, D, B, C
In Bohr’s model, an allowed orbit holds a whole number of de Broglie wavelengths, and the orbit radius grows as \( n^{2} \).
\( \displaystyle 2\pi r_n = n\lambda_n \)
\( \displaystyle \lambda_n = \frac{2\pi r_n}{n} \)
\( \displaystyle r_n \propto n^{2} \;\Rightarrow\; \lambda_n \propto \frac{n^{2}}{n} = n \)
So the de Broglie wavelength increases with n:
\( A\;(n = 1) \lt D\;(n = 2) \lt B\;(n = 3) \lt C\;(n = 4) \)
The increasing order is A, D, B, C.
Shortcut: The electron’s speed in the nth orbit is \( v_n \propto 1/n \), so its momentum falls as 1/n and \( \lambda = h/mv \) rises as n.