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JEE MAIN 2021
27-08-21 S1
Question
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^2}{x m a_0^2}$. The value of $10 x$ is $\_\_\_\_$ ( $\mathrm{a}_0$ is radius of Bohr's orbit)
(Nearest integer) [Given : $\pi=3.14]$
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Question Tags
JEE Main
Chemistry
Easy
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