Question

If $a_0$ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength ($\lambda$) of the electron present in the second orbit of hydrogen atom? [n : any integer]

• A. $\frac{8\pi a_0}{n}$
• B. $\frac{2a_0}{n\pi}$
• C. $\frac{4n}{\pi a_0}$
• D. $\frac{4\pi a_0}{n}$

Answer 1

Answer: (A)

$\frac{8\pi a_0}{n}$

Answer 2

In Bohr’s model, the electron’s de‐Broglie wavelength satisfies the condition

$$n\lambda = 2\pi r_n$$

with

$$r_n = n^2 a_0.$$

Thus,

$$\lambda = \frac{2\pi r_n}{n} = \frac{2\pi n^2 a_0}{n} = 2\pi n a_0.$$

For the second orbit (n = 2):

$$\lambda = 2\pi \times 2 \, a_0 = 4\pi a_0.$$

Since the question asks for a general formula in terms of n, we substitute $\lambda = 2\pi n a_0$ and $n=2$ to get $4\pi a_0$. Now we check the options. Option (1) is written as $\frac{8\pi a_0}{n}$. For n = 2, this equals

$$\frac{8\pi a_0}{2} = 4\pi a_0,$$

which matches our result.