Question

A hypothetical one-electron atom has these energy levels:

E6 = -2 x 10-19 J

E5 = -7 x 10-19 J

E4 = -11 x 10-19 J

E3 = -15 x 10-19 J

E2 = -17 x 10-19 J

E1 = -20 x 10-19 J

(This is not an actual atom that exists.According to the Bohr model, energies should follow the expression discussed in section 6.2 of Silberberg.This exercise prepares you for calculations like Questions 6 and 7 of Problem Set 1.)

(a) If the electron is initially in the n = 4 level, what is the shortest wavelength of radiation that could be emitted?Provide your answer to one significant figure: _______x10-7 m.

(b) What is the ionization energy (in kJ/mol) of the atom in its first excited state?Provide your answer to two significant figures: ______x 103 kJ/mol

Answer 1

Step-by-step explanation:

`E=(h* c)/(\lambda)`

where,

E = energy of photon = ?

h = Planck's constant =
`6.634* 10^(-34)Js`

c = speed of light =
`3* 10^8m/s`

`\lambda` = Wavelength of the radiation

As we can see that wavelength and energy of the radiation are inversely related to each other.So with increase in energy wavelength decreases and vice versa.

Given: The energy of different of different energy levels in an hypothetical model of atom;

`E_6 =-2* 10^(-19) J`

`E_5 = -7* 10^(-19) J`

`E_4 = -11 * 10^(-19) J`

`E_3 = -15* 10^(-19) J`

`E_2 = -17* 10^(-19) J`

`E_1 = -20* 10^(-19) J`

a)If the electron is initially in the n = 4 level. Then maximum difference of energy in energy levels will corresponds to shortest wavelength.

So,
`E=E_4-E_1`

`E=-11 * 10^(-19) J-(-20 * 10^(-19) J)=9 * 10^(-19) J`

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

`=(6.634* 10^(-34)Js* 3* 10^8m/s)/(9 * 10^(-19) J)`

`\lambda = 2.211* 10^(-7)m\approx 2* 10^(-7) m`

`2* 10^(-7) m`is the shortest wavelength of radiation that could be emitted.

b) Ionization energy of the atom in its first excited state.

`E'=E_(\infty )-E_2`

`E'=0-(-17* 10^(-19) J)=17* 10^(-19) J`

Ionization energy of the 1 mole :

`I.E.=(17* 10^(-19) J)/((1)/(6.022* 10^(23)) mol)`

`=1,203,740 J/mol=1203.740 kJ/mol\approx 1.2* 10^3 kJ/mol`

`1.2* 10^3 kJ/mol` is the ionization energy of the atom in its first excited state.