JEE Advance - Chemistry (1984 - No. 2)
Explanation
To calculate the energy required to completely remove an electron from the n = 2 orbit in a hydrogen atom, we need to understand that completely removing an electron refers to ionizing the atom. Ionization requires the electron to move from its given orbit (in this case, n = 2) to essentially n = ∞, where its energy would be 0.
Given, the energy of an electron in a hydrogen atom is:
$ E = \frac{-21.7 \times 10^{-12}}{n^2} $ ergs.
For n = 2 (initial state energy $E_i$):
$ E_i = \frac{-21.7 \times 10^{-12}}{2^2} $
$ E_i = \frac{-21.7 \times 10^{-12}}{4} $
$ E_i = -5.425 \times 10^{-12} $ ergs.
For n = ∞ (final state energy, $E_f$), the energy is 0 since the electron is completely removed from the atom.
The energy required ($E_{required}$) to remove the electron is the difference in energy between the final and initial states:
$ E_{required} = E_f - E_i $
$ E_{required} = 0 - (-5.425 \times 10^{-12}) $
$ E_{required} = 5.425 \times 10^{-12} $ ergs.
To find the longest wavelength ($\lambda_{max}$) of light that can be used to cause this transition (which means providing exactly the energy calculated above), we use the relationship between energy and wavelength provided by the equation:
$ E = \frac{hc}{\lambda} $
Where:
- $E$ is the energy in ergs,
- $h$ is Planck's constant ($6.626 \times 10^{-27}$ erg·s),
- $c$ is the speed of light in vacuum ($3.00 \times 10^{10}$ cm/s),
- $\lambda$ is the wavelength in cm.
Rearranging this equation to solve for $\lambda$:
$ \lambda = \frac{hc}{E} $
Substituting the values:
$ \lambda_{max} = \frac{(6.626 \times 10^{-27}) \times (3.00 \times 10^{10})}{5.425 \times 10^{-12}} $
$ \lambda_{max} = \frac{1.9878 \times 10^{-16}}{5.425 \times 10^{-12}} $
$ \lambda_{max} = 3.664 \times 10^{-5} $ cm, or 366.4 nm.
Thus, the energy required to remove an electron completely from the n = 2 orbit is $5.425 \times 10^{-12}$ ergs, and the longest wavelength of light that can be used to cause this transition is $3.664 \times 10^{-5}$ cm (or 366.4 nm when converted to nanometers for more common wavelength units in light spectra considerations).
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