Question

If $a_0$ be the radius of first Bohr's orbit of H-atom, the de-Broglie's wavelength of an electron revolving in the second Bohr's orbit will be :

• A. $6\pi a_0$
• B. $4\pi a_0$
• C. $2\pi a_0$
• D. None of these

Answer 1

Answer: (B)

$4\pi a_0$

Answer 2

The de Broglie wavelength ($\lambda$) of an electron in a Bohr orbit is related to the orbit's radius ($r$) and the principal quantum number ($n$) by Bohr's second postulate and de Broglie's hypothesis. Combining $mvr = \frac{nh}{2\pi}$ and $\lambda = \frac{h}{mv}$, we get $2\pi r = n\lambda$. The radius of the $n$-th Bohr orbit ($r_n$) is given by $r_n = n^2 r_1$, where $r_1$ is the radius of the first Bohr orbit. In this problem, $r_1 = a_0$. So, for the $n$-th orbit, $r_n = n^2 a_0$. We need to find the de Broglie wavelength for the second Bohr orbit, which means $n=2$. The radius of the second Bohr orbit is $r_2 = 2^2 a_0 = 4a_0$. Using the relation $2\pi r_n = n\lambda_n$, for $n=2$: $2\pi r_2 = 2\lambda_2$, which gives $\lambda_2 = \pi r_2$. Substituting the value of $r_2$: $\lambda_2 = \pi (4a_0) = 4\pi a_0$. Alternatively, we can use the derived formula $\lambda_n = 2\pi r_n / n$. For $n=2$ and $r_2 = 4a_0$, $\lambda_2 = 2\pi (4a_0) / 2 = 4\pi a_0$. Another way is to use the derived formula $\lambda_n = 2\pi r_n / n$. Since $r_n = n^2 a_0$, $\lambda_n = 2\pi (n^2 a_0) / n = 2\pi n a_0$. For $n=2$, $\lambda_2 = 2\pi (2) a_0 = 4\pi a_0$.