How much higher in energy is this electron with respect to the ground state of H

Part A

Suppose an electron is in the n=6 energy level of a H atom.

Part B

How much additional energy is required to just remove this excited electron from the H atom?

Solutions

Expert Solution

1) according bore's formula for energy

E =R_H[(1/n_i²) - (1/n_f²)]

here R_H is constant =2.179 x10^-18 J

n_i =intial level =1 and n_f =final level =6

E =(2.179 x10^-18 J )[(1/1) -(1/36)] = 2.1185 x10^-18 J =2.1185 x10^-18 x(6.241509·10¹⁸ eV) =13.22 eV

energy difference is 13.22 eV

2)additional energy required to remove the electron is ionization energy

E =-Z²R_H (1/n²)

here Z= atomic number of hydrogen =1 and n=6

E =-2.179 x10^-18 J (1/36) =6.0528 x10^-20 J =6.0528 x10^-20 x6.241509·10¹⁸ eV) =0.38 eV

genius_generous answered 1 year ago

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