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Q.

Kinetic and potential energy ( in eV) of electron present in second Bohr’s orbit of hydrogen atom are respectively [Energy of first Bohr’s orbit $= - 13.6 eV]$

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a

$- 1.51, - 3.02$

b

$3.02, - 1.51$

c

$- 13.6, - 27.2$

d

$3.4, - 6.8$

answer is C.

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Detailed Solution

$(KE)_{n} = + 13.6 \frac{Z^{2}}{n^{2}} eV / atom$

$For n = 2, (KE) = + 3.4 eV / atom$

$(PE)_{n} = - 27.2 \frac{Z^{2}}{n^{2}} eV / atom$

$for n = 2, PE = - 6.8 eV / atom$

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Kinetic and potential energy ( in eV) of electron present in second Bohr’s orbit of hydrogen atom are respectively [Energy of first Bohr’s orbit =−13.6eV]