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Annette [7]
3 years ago
11

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
Chemistry
2 answers:
SVETLANKA909090 [29]3 years ago
7 0

. 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.

Furkat [3]3 years ago
3 0

Answer:

The value of ionization energy is 8.716\times 10^{-23} kJ.

Explanation:

E_n=-13.6\times \frac{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\times \frac{1^2}{5^2}eV=-0.544 eV

Z = 1 , n = ∞

E_{\infty }=-13.6\times \frac{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\times 10^{-22} kJ

0.544 eV=0.544\times 1.60218\times 10^{-22} J=8.716\times 10^{-23} kJ

The value of ionization energy is 8.716\times 10^{-23} kJ.

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