De-Broglie wavelength.... | JEE Main 2024

JEE MAINS 2024

09.04.24 S1

Question

The de-Broglie's wavelength of an electron in the $4^{\text {th }}$ orbit is $\_\_\_\_$ $\pi \mathrm{a}_0 \cdot\left(\mathrm{a}_0=\right.$ Bohr's radius $)$

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Solution

$\begin{aligned} & 2 \pi r_n=n \lambda_d \\ & 2 \pi a_0 \frac{n^2}{z}=n \lambda_d \\ & 2 \pi a_0 \frac{4^2}{1}=4 \lambda_d \\ & \lambda_d=8 \pi a_0\end{aligned}$

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JEE Main

Chemistry

Medium

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