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Chemistry · 2021

JEE Main · 27 August 2021, Shift 1 · Q51

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to (h^2)/(x m a_0^2). The value of 10 x is \ \ \ \. ( a_0 is…

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\displaystyle \frac{\mathrm{h}^2}{x \mathrm{~m} \mathrm{a}_0^2}$. The value of $\displaystyle 10 x$ is \_\_\_\_. ( $\displaystyle \mathrm{a}_0$ is radius of Bohr's orbit) (Nearest integer) [ Given : $\displaystyle \pi=3.14$ ]
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.