A) Zero
B) -27.2eV
C) 1 eV
D) 2 eV
Correct Answer: B
Solution :
From the first postulate of Bohrs atom model. \[{{90}^{o}}\] \[{{V}_{R}}\] \[{{V}_{C}}\] i.e., KE of electron \[{{90}^{o}}\] ....(1) Potential due to the nucleus, in the orbit in which electron is revolving \[V=\sqrt{V_{R}^{2}+V_{C}^{2}}\] \[{{V}_{R}}=12V,V=20V\] Potential energy of electron \[\therefore \] \[{{(20)}^{2}}={{(12)}^{2}}+V_{C}^{2}\] ?..(2) Total energy of electron in the orbit \[\Rightarrow \] \[V_{C}^{2}={{(20)}^{2}}-{{(12)}^{2}}\] \[=400-144=256\] ?..(3) This implies that \[\Rightarrow \] Given, that in ground state of hydrogen, total energy \[{{V}_{C}}=\sqrt{256}=16V\] \[F=\frac{IA}{c}\] \[P=\frac{F}{A}=\frac{I}{C}\] \[F\times \Delta t.\] Note: Total energy of electron in outer orbits is more than that in inner orbits.
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