JEE MAIN - Chemistry (2021 - 27th August Morning Shift - No. 15)
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $${{{h^2}} \over {xma_0^2}}$$. The value of 10x is ___________. (a0 is radius of Bohr's orbit) (Nearest integer) [Given : $$\pi$$ = 3.14]
Answer
3155
Explanation
$$mvr = {{nh} \over {2\pi }}$$
$$K.E. = {{{n^2}{h^2}} \over {8{\pi ^2}m{r^2}}} = {{4{h^2}} \over {8{\pi ^2}m{{(4{a_0})}^2}}}$$
$$ = \left( {{4 \over {8{\pi ^2} \times 16}}} \right){{{h^2}} \over {ma_0^2}}$$
$$\Rightarrow$$ x = 315.507
$$\Rightarrow$$ 10x = 3155 (nearest integer)
$$K.E. = {{{n^2}{h^2}} \over {8{\pi ^2}m{r^2}}} = {{4{h^2}} \over {8{\pi ^2}m{{(4{a_0})}^2}}}$$
$$ = \left( {{4 \over {8{\pi ^2} \times 16}}} \right){{{h^2}} \over {ma_0^2}}$$
$$\Rightarrow$$ x = 315.507
$$\Rightarrow$$ 10x = 3155 (nearest integer)
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