Question

An electron is confined to a one dimensional potential box of length 2 Å. Calculate the energies corresponding to the second and fourth quantum states in eV.

Answer 1

Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J

Answer 2

Step-by-Step explanation:

Given: L= 2 Angstrom

To calculate: energies corresponding to quantum states

Particle confined in a one-dimensional box :

For such particles, the energy associated with its quantum number is given by the expression:

$E = \frac{n^{2} h^{2} }{8mL^{2} } - (i)$

where,

n = quantum number

h = planks constant $(6.626 * 10^{-34} J.s)$

m= mass of the particle ( for electron $m= 9.1 * 10^{-31} Kg$

L = length of the box (in meters)

E = energy in Joules

For the second quantum state,

n=2

Substituting the values in equation (i) we get:

$E_{2} = \frac{2^{2} *(6.626*10^{-34} )^{2} }{8*(9.1*10^{-31})^{2} *(2*10^{-10} )^{2} }$

    $= 6.024*10^{-18} J$

Since, $1J= 6.2415*10^{18} eV$

Therefore, $E_{2}= 6.024*10^{-18} *6.2415*10^{18}$

                       $= 37.6 eV$

Similarly, we calculate Energy for the fourth quantum state

In this case, n= 4

Solving we get, $E_{4} = 150.412 eV$

Hence, The energies corresponding to the second and fourth quantum states in $37.6 eV$ and $150.412 eV$