Question

An electron confined in a one-dimensional box is observed, at different times, to have energies of 12 eV, 27 eV, and 48 eV. What is the length of the box? 45. | The nucleus of a typical atom

Answer 1

Answer:

$l=3.5*10^{-10}m$

Explanation:

From the question we are told that:

1st Energy     $E_1=12eV=4(3eV)$

2nd Energy  $E_2=27eV=9(3eV)$

3rd Energy   $E_3=48eV=16(3eV)$

 

Generally the equation for Energy E for electron in one dimensional box at ground state $E_0$ is mathematically given by

  $E_0=\frac{h}{8ml^2}$

  $E_0=\frac{h}{8ml^2}$

Therefore Length a is mathematically given as

$l=\sqrt{\frac{h^2}{8mE_0} }$

$l=\sqrt{\frac{(6.625*10^{-34})^2}{8(9.1*19^{-31}{(3eV(1.6*10^{-19}))}}$

$l=3.5*10^{-10}m$