The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h ^ 2 x ma _ 0 ^ 2 . The value of 10x is ( a _ 0 is radius of Bohr's orbit) (Nearest integer) [Given: π = 3 . 14 ]

Correct answer: 0 x = 8 π ^ 2 × 16 4 = 32 π ^ 2 = 32 ( 3 . 14 ) ^ 2 = 315 . 5072 10x = 315 . 5072

JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2021 · 27 August · Morning Shift · Chemistry · Chemical Kinetics

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 $\dfrac{{h}^{2}}{x{ma}_{0}^{2}} .$ The value of $10x$ is ( ${a}_{0}$ is radius of Bohr's orbit)

(Nearest integer)

[Given: $\pi = 3 . 14$ ]

Correct answer

Step-by-step solution

$x = \dfrac{8{\pi}^{2} \times 16}{4} = 32{\pi}^{2}$

$= 32 ( 3 . 14{)}^{2}$

$= 315 . 5072$

$10x = 315 . 5072$

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