The de Broglie wavelength of the electron in | vedclass

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(D) According to the Bohr model, the radius of the $n^{th}$ orbit is given by $r_n = n^2 a_0$, where $a_0$ is the Bohr radius.The circumference of the

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$3\lambda_1$

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(D) According to the Bohr model, the radius of the $n^{th}$ orbit is given by $r_n = n^2 a_0$, where $a_0$ is the Bohr radius.The circumference of the $n^{th}$ orbit is $2\pi r_n = 2\pi n^2 a_0$.According to the de Broglie hypothesis, the condition for a stable orbit is $n\lambda = 2\pi r_n$.Substituting the expression for $r_n$, we get $n\lambda_n = 2\pi n^2 a_0$, which simplifies to $\lambda_n = 2\pi n a_0$.For the ground state $(n = 1)$, $\lambda_1 = 2\pi(1)a_0 = 2\pi a_0$.For the $n = 3$ level, $\lambda_3 = 2\pi(3)a_0 = 3(2\pi a_0) = 3\lambda_1$.Therefore, $\lambda_3 = 3\lambda_1$.

The de Broglie wavelength of the electron in the ground state is $\lambda_1$ and that in the $n = 3$ level is $\lambda_3$. Then $\lambda_3$ is given by:

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