Question

Calculate the energy required to ionize a hydrogen atom to an excited state where the electron is initially in the n = 5 energy level. Report your answer in kilojoules

Answer 1

. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:

E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ

Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ

2. The wavelength of a photon with this energy would be:

Energy = hc/wavelength

wavelength = hc/energy

wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m

Converting to nanometers gives: 91.16 nm

3. Repeat the calculation in 1, but using n=5.

4. Repeat the calculation in 2 using the energy calculated in 3.

Answer 2

The value of ionization energy is
`8.716* 10^(-23) kJ`.

Step-by-step explanation:

`E_n=-13.6* (Z^2)/(n^2)eV`

where,

`E_n` = energy of
`n^(th)` orbit

n = number of orbit

Z = atomic number

We are given: Z = 1 , n= 5

`E_5=-13.6* (1^2)/(5^2)eV=-0.544 eV`

Z = 1 , n = ∞

`E_(\infty )=-13.6* (1^2)/((\infty )^2)}eV=0 eV`

Ionization energy to ionize the hydrogen atom where electron is present in n=5 initially will equal to the difference between energy of electron at infinity from the energy of electron at n = 5 energy level.

`I.E=E_(\infty )-E_5`

`=0 eV-(-0.544 eV)=0.544 eV`

`1 eV= 1.60218* 10^(-22) kJ`

`0.544 eV=0.544* 1.60218* 10^(-22) J=8.716* 10^(-23) kJ`

The value of ionization energy is
`8.716* 10^(-23) kJ`.