The kinetic energy of an electron in an orbit of hydrogen atom is 3.4ev/atom. Then identify the correctly matched set for that electron
List -1 List -2
A)Potential energy 1) 1.09×10⁸cm/sec
B) Total energy 2) 2.116×10^–8 cm
C) Velocity 3) –6.8 ev/atom
D) Its distance from nucleous 4) –3.4 ev/atom
The correct match is

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A B C D

4 3 2 1

A B C D

3 4 2 1

A B C D

3 4 1 2

A B C D

2 1 4 3

answer is B.

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Detailed Solution

kinetic energy of an electron in an orbit of hydrogen atom is 3.4ev/atom.

TE=-KE

Hence TE= -3.4

$E = - \frac{13.6}{n^{2}} - 3.4 = - \frac{13.6}{n^{2}} n = 2$

Distance from the nucleous (r)

$r = 0.529 \times n^{2} \times 10^{- 8} c m = 0.529 \times 2^{2} \times 10^{- 8} c m = 2.11 \times 10^{- 8} c m$

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