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How much energy is required to ionise an H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).

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#### Solution

The expression of energy is given by,

`"E"_"n" = (-(2.18xx10^(-18))"Z"^2)/"n"^2`

Where,

Z = atomic number of the atom

n = principal quantum number

For ionization from n_{1} = 5 to `"n"_2 = ∞`

`triangle "E" = "E"_(∞) - "E"_5`

`=[{{-(2.18xx10^(-18)J)(1)^2)/(∞)^2}-{(-(2.18xx10^(-18)J(1)^2))/(5)^2}]`

`= (2.18xx10^(-18)"J")(1/((5)^2)) ("Since" 1/∞ = 0)`

`= 0.0872 xx 10^(-18) "J"`

`triangle "E" = 8.72 xx 10^(-20) "J"`

Hence, the energy required for ionization from n = 5 to n = ∞ is 8.72 × 10^{–20} J.

Energy required for n_{1} = 1 to n = ∞

`triangle "E'" = "E"_(∞) - "E"_1`

`= [{(-(2.18xx10^(-18))(1)^2)/(∞)^2}-{(-(2.18xx10^(-18))(1)^2)/(1)^2}]`

`= (2.18 xx 10^(-18)) [1 - 0]`

`= 2.18 xx 10^(-18) "J"`

Hence, less energy is required to ionize an electron in the 5^{th} orbital of hydrogen atom as compared to that in the ground state.

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