Question

Calculate the energy difference between the two energy levels if an electron jumps from 2nd excited State to 5th excited state and the length of the box is 0.760nm

Answer 1

Given that,

Length = 0.760 nm

The two energy levels if an electron jumps from 2nd excited State to 5th excited state.

We know that,

The energy is

$E=\dfrac{n^2\pi^2(\hbar)^2}{2ma^2}$

For ground state,

n = 1

For second excited state,

n = 3

For fifth excited state,

n = 6

If an electron jumps from 2nd excited State to 5th excited state.

We need to calculate the energy difference between the two energy levels

Using formula of energy

$\Delta E=E_{5}-E_{2}$

$\Delta E=\dfrac{n^2\pi^2\times(\hbar)^2}{2ma^2}-\dfrac{n^2\pi^2(\hbar)^2}{2ma^2}$

$\Delta E=\dfrac{\pi^2\times(\hbar)^2}{2ma^2}(n_{5}^2-n_{2}^2)$

Put the value into the formula

$\Delta E=\dfrac{\pi^2\times h^2}{4\pi^2\times2ma^2}(36-9)$

$\Delta E=\dfrac{(6.67\times10^{-34})^2}{8\times9.1\times10^{-31}\times(0.760\times10^{-9})^2}\times27$

$\Delta E=28.5\times10^{-19}$

$\Delta E=\dfrac{28.5\times10^{-19}}{1.6\times10^{-19}}$

$\Delta E=17.81\ eV$

Hence, The energy difference between the two energy levels is 17.81 eV.