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Example · Example 2

Q.Calculate the energy of an electron in the second Bohr orbit of a hydrogen atom. (Given: En=−2.18×10−18 Z2n2 JE_n = -2.18 \times 10^{-18}\ \dfrac{Z^2}{n^2}\ \text{J}.)

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✓ Free question

For a hydrogen atom, Z=1Z = 1. The energy of the electron in the nthn^{\text{th}} Bohr orbit is

En=−2.18×10−18 Z2n2 JE_n = -2.18 \times 10^{-18}\,\frac{Z^2}{n^2}\ \text{J}

For the second orbit, n=2n = 2:

E2=−2.18×10−18×1222=−2.18×10−18×14=−5.45×10−19 JE_2 = -2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -2.18 \times 10^{-18} \times \frac{1}{4} = -5.45 \times 10^{-19}\ \text{J}

The negative sign shows the electron is bound to the nucleus (its energy is lower than that of a free, stationary electron infinitely far from the nucleus, which is taken as the zero of energy).

✓Final answer

The energy of the electron in the second Bohr orbit of hydrogen is −5.45×10−19 J-5.45 \times 10^{-19}\ \text{J}.

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