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QJEEM21Chd678: JEE Main 2021 Chemistry - Class 11 JEEM21Chd678 Solution: The kinetic energy of an electron in the second Bohr orbit of a …

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h2xma02{{{h^2}} \over {xma_0^2}} . The value of 10x is ___________. (a0 is radius of Bohr's orbit) (Nearest integer) [Given : π\pi = 3.14]

Step-by-Step Verified Solution

mvr=nh2πmvr = {{nh} \over {2\pi }}

K.E.=n2h28π2mr2=4h28π2m(4a0)2K.E. = {{{n^2}{h^2}} \over {8{\pi ^2}m{r^2}}} = {{4{h^2}} \over {8{\pi ^2}m{{(4{a_0})}^2}}}

=(48π2×16)h2ma02= \left( {{4 \over {8{\pi ^2} \times 16}}} \right){{{h^2}} \over {ma_0^2}}

⇒\Rightarrow x = 315.507

⇒\Rightarrow 10x = 3155 (nearest integer)

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