Physics of Everyday Phenomena
Physics of Everyday Phenomena
9th Edition
ISBN: 9781259894008
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
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Chapter 18, Problem 3SP

(a)

To determine

The energy required to ionize a hydrogen atom in its lowest energy level.

(a)

Expert Solution
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Answer to Problem 3SP

The energy required to ionize a hydrogen atom in its lowest energy level is 13.6 eV.

Explanation of Solution

The energy level diagram of hydrogen proposed by the Bohr model gives the energy of electrons in different orbits in hydrogen atom. The energy of the ground state or the lowest level in the hydrogen atom is 13.6 eV.

Ionization energy is the energy required to remove an electron from its orbit. It will be equal to the negative of the energy of the electron in the orbit. The energy of the electron in the lowest level of hydrogen atom is 13.6 eV. The negative of 13.6 eV is 13.6 eV.

Conclusion:

Thus the energy required to ionize a hydrogen atom in its lowest energy level is 13.6 eV.

(b)

To determine

The energy required to ionize the hydrogen atom when it is in the first excited state above the lowest level.

(b)

Expert Solution
Check Mark

Answer to Problem 3SP

The energy required to ionize the hydrogen atom when it is in the first excited state above the lowest level is 3.4 eV.

Explanation of Solution

The energy level diagram of hydrogen proposed by the Bohr model gives the energy of electrons in different orbits in hydrogen atom. The energy of a particular level in the hydrogen atom is given by the ratio of 13.6 eV to the square of the principal quantum number of the particular energy level.

The first excited state is the level which lies just above the ground level. The energy of the first excited state of hydrogen atom is 3.4 eV.

Ionization energy is the energy required to remove an electron from its orbit. It will be equal to the negative of the energy of the electron in the orbit. This implies the energy required to ionize the hydrogen atom in the first excited state will be equal to 3.4 eV.

Conclusion:

Thus the energy required to ionize the hydrogen atom when it is in the first excited state above the lowest level is 3.4 eV.

(c)

To determine

The wavelength of the photon emitted when an ionized hydrogen atom goes to the lowest energy level after capturing an electron with zero kinetic energy.

(c)

Expert Solution
Check Mark

Answer to Problem 3SP

The wavelength of the photon emitted when an ionized hydrogen atom goes to the lowest energy level after capturing an electron with zero kinetic energy is 91.3 nm.

Explanation of Solution

An ionized hydrogen atom will have zero energy. The energy of the lowest level in the hydrogen atom is 13.6 eV. The energy of the photon emitted will be equal to the energy difference between the final and initial states of the atom.

Write the equation for the energy of the photon.

E=EfEi

Here,

E is the energy of the photon

Ef is the final energy of the atom

Ei is the initial energy of the atom

Substitute 0 eV for Ef and 13.6 eV for Ei in the above equation to find E.

E=0 eV(13.6 eV)=13.6 eV

Write the equation for the energy of a photon.

E=hcλ

Here,

h is the Planck’s constant

c is the speed of light in vacuum

Rewrite the above equation for λ.

λ=hcE

The value of h is 6.626×1034 Js and the value of c is 3.0×108 m/s.

Substitute 6.626×1034 Js for h,3.0×108 m/s for c and 13.6 eV for E in the above equation to find λ.

λ=(6.626×1034 Js)(3.0×108 m/s)13.6 eV(1.60×1019 J1 eV)=(6.626×1034 Js)(3.0×108 m/s)2.176×1018 J=91.3×109 m=91.3 nm

Conclusion:

Thus the wavelength of the photon emitted when an ionized hydrogen atom goes to the lowest energy level after capturing an electron with zero kinetic energy is 91.3 nm

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