Question

An electron in a box is in the ground state with energy 2.0 eV. (a) Find the width of the box. (b) How much energy is needed to excite the electron to its first excited state? (c) If the electron makes a transition from an excited state to the ground state with the simultaneous emission of 30.0-eV photon, find the quantum number of the excited state?

Answer 1

Explanation:

Energy of the electron in the ground state, $E_o=2\ eV=2\times 1.6\times 10^{-19}\ J=3.2\times 10^{-19}\ J$

(a) The energy of the electron is square well is given by :

$E=\dfrac{n^2h^2}{8ml}$  

Where

l is the width of the box

$E_o=\dfrac{n^2h^2}{8ml}$  

$l=\dfrac{n^2h^2}{8mE_o}$  

$l=\dfrac{(1)^2\times (6.63\times 10^{-34})^2}{8\times 9.1\times 10^{-31}\times 3.2\times 10^{-19}}$

$l=1.88\times10^{-19}\ m$

(b) For first excited state, n = 2

$E_1=\dfrac{(n)^2h^2}{8ml}$  

$E_1=\dfrac{(2)^2\times ((6.63\times 10^{-34})^2)^2}{8\times 9.1\times 10^{-31}\times 1.88\times10^{-19}}$  

$E_1=1.28\times 10^{-18}\ J$

(c) Let n is the value of quantum number of the excited state. Again using this formula as :

$E=\dfrac{(n)^2h^2}{8ml}$  

$n=\sqrt{\dfrac{8ml E}{h^2}}$

$n=\sqrt{\dfrac{8 \times 9.1\times 10^{-31}\times 1.88 \times 10^{-19} \times 30\times 1.6\times 10^{-19}}{(6.63\times 10^{-34})^2}}$

n = 3.86

or

n = 4

Hence, this is the required solution.