The total energy of an electron in the first excited state of the hydrogen atom is -3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state.

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Solution

Kinetic energy of electron is, $K E = \frac{13.6 Z^{2}}{n^{2}} e V$
For the first excited state of the hydrogen atom, $n = 2$ and $Z = 1$
$∴ K E = \frac{13.6}{2^{2}} = 3.4 e V$
Total energy , $E = K E + P E \Rightarrow - 3.4 = 3.4 + P E$
$∴ P E = - 6.8 e V$

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The total energy of an electron in the first excited state of the hydrogen atom is -3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state.

Kinetic energy of electron is, KE=13.6Z2n2eVFor the first excited state of the hydrogen atom, n=2 and Z=1∴KE=13.622=3.4eVTotal energy , E=KE+PE⇒−3.4=3.4+PE∴PE=−6.8 eV

Kinetic energy of electron is, $K E = \frac{13.6 Z^{2}}{n^{2}} e V$
For the first excited state of the hydrogen atom, $n = 2$ and $Z = 1$
$∴ K E = \frac{13.6}{2^{2}} = 3.4 e V$
Total energy , $E = K E + P E \Rightarrow - 3.4 = 3.4 + P E$
$∴ P E = - 6.8 e V$