A) \[\frac{13.6}{{{n}^{4}}}eV\]
B) \[\frac{13.6}{{{n}^{3}}}eV\]
C) \[-\frac{13.6}{{{n}^{2}}}eV\]
D) \[\frac{13.6}{n}eV\]
Correct Answer: C
Solution :
Energy (E) of the electron in various Bohr?s orbits: It may be calculated from the folk formula. \[{{E}_{n}}=-\frac{2{{\pi }^{2}}m{{e}^{4}}{{Z}^{2}}}{{{n}^{2}}{{h}^{2}}}\] (in CGS system) Where, e = charge of electron m = mass of electron n = principal quantum number 1, 2, 3 h = Planck's constant Z = number of proton present Value of Z for H-atom is 1 \[{{E}_{n}}=\frac{-13.6{{Z}^{2}}}{{{n}^{2}}}eV/atom\] \[{{E}_{n}}=\frac{-13.6\times {{(1)}^{2}}}{{{n}^{2}}}\] \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV/atom\] Negative sign show that energy of electron increases with the increase in the value of n.
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