The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is \(x × 10^{–8}\ m\). The value of \(x\) is ______. (Nearest Integer) (Given : Energy of the electron in the first shell of the hydrogen atom is \(2.2 \times 10^{–18} J\); \(h = 6.63 × 10^{–34}Js\) and \(c = 3 × 10^8 ms^{–1}\))
The ionization energy of lithium in its ground state can be determined by using the formula \(E = \dfrac{-13.6 \times Z^2}{n^2}\). For lithium (\(Z = 3\)), in its ground state (\(n = 1\)): \(E = \dfrac{-13.6 \times 3^2}{1^2} = -122.4 \ eV\). Convert this energy to joules by multiplying with \(1.6 \times 10^{-19} J/eV\): \(E = -122.4 \times 1.6 \times 10^{-19}\ J = -1.9584 \times 10^{-17}\ J\). The energy of a photon is given by \(E_\text{photon} = \dfrac{hc}{\lambda}\). Thus, \(\lambda = \dfrac{hc}{E}\). Substitute the known values (energy is positive as we consider absolute values): \(\lambda = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.9584 \times 10^{-17}}\ m \approx 1.01 \times 10^{-7}\ m\). Express this wavelength as \(x \times 10^{-8}\ m\): \(\lambda = 10.1 \times 10^{-8}\ m\). Hence, \(x = 10.1\) which rounds to the nearest integer, confirmed as 10. Validate that it falls within the range: 10 (4,4), so \(x = 10\) is indeed correct.
