
The electron energy in a hydrogen atom is given by $ {E_n} = ( - 2.18 \times {10^{ - 18}}){n^2}joules $ . Calculate the energy required to remove an electron completely from the $ n = 2orbit $ . What is the longest wavelength (in A) of light that can be used to cause this transition?
Answer
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Hint: The structure of the atom was a much-debated topic in the mid-1920s. Numerous atomic models including the theory proposed by J.J Thompson and the discovery of the nucleus by Ernest Rutherford had emerged. But it was Neil Bohr, who asserted that electrons revolved around a positively charged nucleus just like the planets around the sun. In the Bohr model of the hydrogen atom, an assumption concerning the quantization of atoms was made which stated that electrons orbited the nucleus in specific orbits or shells with a fixed radius. Only those shells with a radius provided by the equation below were allowed, and it was impossible for electrons to exist between these shells.
Complete step by step solution:
$ {E_n} = ( - 2.18 \times {10^{ - 18}}){n^2}joules $
The energy required to remove an electron completely from $ n = 2orbit $
$ {E_2} = ( - 2.18 \times {10^{ - 18}}){2^2} = - 8.72 \times {10^{ - 18}} $
$ 8.72 \times {10^{ - 18}} = \dfrac{{hc}}{\lambda } $
$ \lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{8.72 \times {{10}^{ - 18}}}} = 2.27958 \times {10^{ - 8}}m = 227.96{A^0} $ .
Note:
An electron absorbs energy in the form of photons and gets excited to a higher energy level. After jumping to the higher energy level, also called the excited state, the excited electron is less stable, and therefore, would quickly emit a photon to come back to a lower and more stable energy level. The emitted energy is equivalent to the difference in energy between the two energy levels for a specific transition. The energy of an electron depends on its location with respect to the nucleus of an atom.
Complete step by step solution:
$ {E_n} = ( - 2.18 \times {10^{ - 18}}){n^2}joules $
The energy required to remove an electron completely from $ n = 2orbit $
$ {E_2} = ( - 2.18 \times {10^{ - 18}}){2^2} = - 8.72 \times {10^{ - 18}} $
$ 8.72 \times {10^{ - 18}} = \dfrac{{hc}}{\lambda } $
$ \lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{8.72 \times {{10}^{ - 18}}}} = 2.27958 \times {10^{ - 8}}m = 227.96{A^0} $ .
Note:
An electron absorbs energy in the form of photons and gets excited to a higher energy level. After jumping to the higher energy level, also called the excited state, the excited electron is less stable, and therefore, would quickly emit a photon to come back to a lower and more stable energy level. The emitted energy is equivalent to the difference in energy between the two energy levels for a specific transition. The energy of an electron depends on its location with respect to the nucleus of an atom.
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