Question

Consider the following energy levels of a hypothetical atom:E4 −1.21 × 10−19 J E3 −5.71 × 10−19 J E2 −1.05 × 10−18 J E1 −1.55 × 10−18 J (a) What is the wavelength of the photon needed to excite an electron from E1 to E4?(b) What is the energy (in joules) a photon must have in order to excite an electron from E2 to E3?(c) When an electron drops from the E3 level to the E1 level, the atom is said to undergo emission. Calculate the wavelength of the photon emitted in this process.

Answer 1

Step-by-step explanation:

Energy levels of a hypothetical atom:

`E_4 =-1.21* 10^(-19) J`

`E_3 =-5.71* 10^(-19) J`

`E_2 =-1.05* 10^(-18) J`

`E_1 =-1.55* 10^(-18) J`

a) Wavelength of the photon needed to excite an electron from
`E_1` to
`E_4`.

Energy difference between forth and first energy level =

`E=E_4-E_1=(-1.21* 10^(-19) J )-(-1.55* 10^(-18) J)=1.429* 10^(-18) J`

`E=(hc)/(\lambda )`

`\lambda =(hc)/(E)=((6.634* 10^(-34)Js)* (3* 10^8m/s))/(1.429* 10^(-18) J)`

`=1.392* 10^(-7) m=139.2 nm`

139.2 nm is the wavelength of the photon needed to excite an electron from
`E_1` to
`E_4`.

b) Energy of a photon in order to excite an electron from
`E_2` to
`E_3`.

Energy difference between third and second energy level =

`E=E_3-E_2=(-5.71* 10^(-19) J)-(-1.05* 10^(-18) J)=4.79* 10^(-19) J`

`4.79* 10^(-19) J` is the energy a photon to excite an electron from
`E_2` to
`E_3`.

c) Electron drops from the
`E_3` level to the
`E_1` level

Energy difference between third and first energy level =

`E=E_3-E_1=(-5.71* 10^(-19) J )-(-1.55* 10^(-18) J)=9.79* 10^(-19) J`

`E=(hc)/(\lambda )`

`\lambda =(hc)/(E)=((6.634* 10^(-34)Js)* (3* 10^8m/s))/(9.79* 10^(-19) J)`

`=2.033* 10^(-7) m=203.3 nm`

202.3 nm is the wavelength of the photon of emitted.