(a) Explanation Given info: The ionization energy of the atom is 17.50 eV . The minimum energy to ionize an atom is when the energy of the atom in the upper state is zero. Therefore the energy of the first level is − 17.50 eV . For transition from fifth level to first level the wavelength is given by 73.86 nm . The energy difference between the fifth level and first level is E 5 − E 1 = h c λ (1) Substitute 1.240 × 10 − 6 eVm for h c and 73.86 nm for wavelength in equation (1) to get the energy. E 5 − E 1 = 1.240 × 10 − 6 eVm 73.8 × 10 − 9 m = 16.79 eV Therefore, E 5 = 16.79 + E 1 = 16.79 − 17.50 = − .71 eV For transition from fourth level to first level the wavelength is given by 75.63 nm . The energy difference between the fourth level and first level is E 4 − E 1 = h c λ (2) Substitute 1.240 × 10 − 6 eVm for h c and 75.63 nm for wavelength in equation (2) to get the energy. E 4 − E 1 = 1.240 × 10 − 6 eVm 75.63 × 10 − 9 m = 16.4 eV Therefore, E 4 = 16.4 + E 1 = 16.4 − 17.50 = − 1.10 eV For transition from third level to first level the wavelength is given by 73.86 nm . The energy difference between the third level and first level is E 3 − E 1 = h c λ (3) Substitute 1 (b) To determine Wavelength of light emitted when transition occurs from fourth to second level.

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Chapter 39, Problem 39.27E

(a)

To determine

The energy of atom in the first, second level etc.

(b)

To determine

Wavelength of light emitted when transition occurs from fourth to second level.

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Chapter 39, Problem 39.26E

Chapter 39, Problem 39.28E

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Chapter 39 Solutions

University Physics with Modern Physics, Books a la Carte Edition; Modified MasteringPhysics with Pearson eText -- ValuePack Access Card -- for ... eText -- Valuepack Access Card (14th Edition)

Ch. 39.2 - Prob. 39.2TYU Ch. 39.3 - Prob. 39.3TYU Ch. 39.4 - Prob. 39.4TYU Ch. 39.5 - Prob. 39.5TYU Ch. 39.6 - Prob. 39.6TYU Ch. 39 - Prob. 39.1DQ Ch. 39 - Prob. 39.2DQ Ch. 39 - Prob. 39.3DQ Ch. 39 - When an electron beam goes through a very small...Ch. 39 - Prob. 39.5DQ

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The energy of atom in the first, second level etc.

Solution Summary: The author explains that the ionization energy of an atom is 17.50eV.

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The energy of atom in the first, second level etc.

Wavelength of light emitted when transition occurs from fourth to second level.

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Chapter 39 Solutions