Calculate the maximum wavelength of light capable of removing an electron for a hydrogen atom from the energy state characterized by \(n=1,\) by \(n=2\)

To find the energy of an electron in a hydrogen atom, we will use the formula: \[E_n = -\frac{13.6 \, eV}{n^2}\] where \(E_n\) is the energy of the electron in the nth energy level, and n is the principal quantum number. For \(n=1\), calculate the energy: \[E_1 = -\frac{13.6 \, eV}{1^2} = -13.6 \, eV\] For \(n=2\), calculate the energy: \[E_2 = -\frac{13.6 \, eV}{2^2} = -\frac{13.6 \, eV}{4} = -3.4 \, eV\]

When an electron is completely removed from an atom, it goes to the energy state with an energy of 0 eV. To find the energy required to remove the electron from each energy state, we have to calculate the difference between the energies in that state and the energy at 0 eV. Energy required to remove the electron from the state \(n=1\): \[\Delta E_1 = E_{\infty} - E_1 = 0 \, eV - (-13.6 \, eV) = 13.6 \, eV\] Energy required to remove the electron from the state \(n=2\): \[\Delta E_2 = E_{\infty} - E_2 = 0 \, eV - (-3.4 \, eV) = 3.4 \, eV\]

The energy and wavelength of a photon are related through the equation: \[E = h\cdot c / \lambda\] where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(4.135667696 \times 10^{-15} eV\cdot s\)), \(c\) is the speed of light (\(3 \times 10^8 m/s\)), and \(\lambda\) is the wavelength of the photon. Rearrange the equation to find the wavelength that corresponds to the energy: \[\lambda = \frac{h\cdot c}{E}\] For the energy state n=1: \[\lambda_1 = \frac{4.135667696 \times 10^{-15} eV\cdot s \cdot 3 \times 10^8 m/s}{13.6 \, eV} = 91.18 \times 10^{-9} m\] For the energy state n=2: \[\lambda_2 = \frac{4.135667696 \times 10^{-15} eV \cdot s \cdot 3 \times 10^8 m/s}{3.4 \, eV} = 364.74 \times 10^{-9} m\] So, the maximum wavelength of light capable of removing an electron for a hydrogen atom from the energy state characterized by \(n=1\) is approximately 91.18 nm, and for the energy state characterized by \(n=2\) is approximately 364.74 nm.