Question 274: Dual Nature of Matter and Radiation

Question

When the electron orbiting in hydrogen atom goes from one orbit to another orbit (principal quantum number $=n$ ), the de-Broglie wavelength ( $\lambda$ ) associated with it is related to $n$ as

Accepted Answer

Step 1. Recall BohrÃ¢ÂÂs quantization condition
BohrÃ¢ÂÂs model says:
$ m v r = n \hbar \quad \quad (1) $
And also, de BroglieÃ¢ÂÂs condition:
$ 2 \pi r = n \lambda \quad \quad (2) $
Thus, the circumference of the orbit equals an integer multiple of the de Broglie wavelength.
Step 2. BohrÃ¢ÂÂs orbit radius
For a hydrogen atom,
$ r_n = a_0 n^2, $
where $a_0$ is the Bohr radius.
Step 3. Relation between ÃÂ» and n
From equation (2):
$ \lambda = \frac{2 \pi r_n}{n}. $
Substitute $r_n = a_0 n^2$:
$ \lambda = \frac{2 \pi (a_0 n^2)}{n} = 2 \pi a_0 n. $
Step 4. Conclusion
So, the de Broglie wavelength is
directly proportional to $n$
.
$ \boxed{\lambda \propto n} $
Correct Option: D
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Answer

$\lambda \propto \frac{1}{\mathrm{n}^2}$

Answer

$\lambda \propto n^2$

Answer

$\lambda \propto \frac{1}{n}$

Answer

$\lambda \propto \mathrm{n}$

Dual Nature of Matter and Radiation

When the electron orbiting in hydrogen atom goes from one orbit to another orbit (principal quantum number =n ), the de-Broglie wavelength ( \lambda ) associated with it is related to n as

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