Question #141997

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

Expert's answer

"\\begin{aligned}\nL &= 1nm = 10^{-9}m\\\\\nn &= 1 \\ \\textsf{(ground state)}\\\\\nn &= 2 \\ \\textsf{(first excited state)}\\\\\nm_e\\ \\textsf{(m. of electron)} &= 9.11\u00d7 10^{-31} kg\\\\\nh &= 6.63 \u00d7 10^{-34} Js\\\\\nE_n &= \\dfrac{n^2h^2}{8m_eL^2}\n\\end{aligned}"

"\\begin{aligned}\n\nE_1 &= \\dfrac{n^2h^2}{8m_eL^2}\\\\\n\\\\\nE_1 &= \\dfrac{1^2\u00d7{(6.63\u00d710^{-34})^2}}{8(9.11\u00d710^{-31}) \u00d7 (10^{-9})^2}\\\\\n\\\\\n&= 6.031\u00d710^{-20}J\\\\\n&= \\dfrac{6.031\u00d710^{-20}}{1.6\u00d710^{-19}}\\ eV\\\\\n\\\\\n&= 0.377eV\n\\end{aligned}"

"\\begin{aligned}\n\nE_1 &= \\dfrac{n^2h^2}{8m_eL^2}\\\\\n\\\\\nE_1 &= \\dfrac{2^2\u00d7{(6.63\u00d710^{-34})^2}}{8(9.11\u00d710^{-31}) \u00d7 (10^{-9})^2}\\\\\n\\\\\n&= 4(0.377eV)\\\\\n&= 1.508eV\n\\end{aligned}"

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

Learn more about our help with Assignments: Inorganic Chemistry

## Comments

## Leave a comment