Answer to Question #141997 in Inorganic Chemistry for Sarita bartwal

Question

Calculate the energy required to excite an electron in a one-dimensional box from its ground state to first excited state. Given that l=1nm

Answer 1

$\begin{aligned} L &= 1nm = 10^{-9}m\\ n &= 1 \ \textsf{(ground state)}\\ n &= 2 \ \textsf{(first excited state)}\\ m_e\ \textsf{(m. of electron)} &= 9.11× 10^{-31} kg\\ h &= 6.63 × 10^{-34} Js\\ E_n &= \dfrac{n^2h^2}{8m_eL^2} \end{aligned}$

$\begin{aligned} E_1 &= \dfrac{n^2h^2}{8m_eL^2}\\ \\ E_1 &= \dfrac{1^2×{(6.63×10^{-34})^2}}{8(9.11×10^{-31}) × (10^{-9})^2}\\ \\ &= 6.031×10^{-20}J\\ &= \dfrac{6.031×10^{-20}}{1.6×10^{-19}}\ eV\\ \\ &= 0.377eV \end{aligned}$

$\begin{aligned} E_1 &= \dfrac{n^2h^2}{8m_eL^2}\\ \\ E_1 &= \dfrac{2^2×{(6.63×10^{-34})^2}}{8(9.11×10^{-31}) × (10^{-9})^2}\\ \\ &= 4(0.377eV)\\ &= 1.508eV \end{aligned}$

$\therefore$ The energy required to excite the electron = 1.508eV - 0.377eV =1.131eV