Question

An electron is trapped in an infinite square-well potential of width 0.4 nm. If the electron is initially in the n = 4 state, what are the various photon energies that can be emitted as the electron jumps to the ground state? (List in descending order of energy. Enter 0 in any remaining unused boxes.)answers are, but cannot solve to ithighest: 35.3 eV28.2 eV18.8 eV16.5 eV11.8 eVlowest: 7.06 eV

Answer 1

The energies in descending order 35.3 eV, 28.2 eV, 18.8 eV, 16.5 eV, 11.8 eV, 7.05 eV.

Step-by-step explanation:

Given that,

Width = 0.4 nm

Number of state n= 4

We need to calculate the energy in first state

When n = 1,

Using formula of energy

`E_(1)=(h^2)/(8mL^2)`

Put the value into the formula

`E_(1)=((6.63*10^(-34))^2)/(8*9.1*10^(-31)*(0.4*10^(-9))^2)`

`E_(1)=3.77*10^(-19)\ J`

`E_(1)=(3.77*10^(-19))/(1.6*10^(-19))`

`E_(1)=2.35\ eV`

Energy in nth level is

`E_(n)=n^2* E_(1)`

Photon energy in different states is

`E_(p)=E_(n)-E_(n-1)`

Put the value into the formula

`E_(4-3)=(4^2-3^2)*2.35=16.5\ eV`

`E_(4-2)=(4^2-2^2)*2.35=28.2\ eV`

`E_(4-1)=(4^2-1^2)*2.35=35.3\ eV`

`E_(3-2)=(3^2-2^2)*2.35=11.8\ eV`

`E_(3-1)=(3^2-1^2)*2.35=18.8\ eV`

`E_(2-1)=(2^2-1^2)*2.35=7.05\ eV`

Hence, The energies in descending order 35.3 eV, 28.2 eV, 18.8 eV, 16.5 eV, 11.8 eV, 7.05 eV.