a. Explanation Given data: Wavelengths of absorption spectrum is 200 nm, 300 nm and 500 nm. Formula used: Write the expression for energy. E photon = h c λ (1) Here, E photon is the energy, h is the Planck’s constant, which has the standard value as 6.62 × 10 − 34 m 2 ⋅ kg s , c is the speed of the light 3 × 10 8 m s , and λ is the wavelength. Explanation: For 200 nm, substitute 6.62 × 10 − 34 m 2 ⋅ kg s for h , 3 × 10 8 m s for c and 200 × 10 − 9 m for λ in equation (1), E photon = ( 6.62 × 10 − 34 m 2 ⋅ kg s ) × ( 3 × 10 8 m s ) 200 × 10 − 9 m = ( 1.986 × 10 − 25 200 × 10 − 9 ) J = 9.93 × 10 − 19 J Consider the conversion factor of J to eV. 1 J = ( 1 1.6 × 10 − 19 ) eV Consider the expression of energy in eV. E photon = 9.93 × 10 − 19 J 1.6 × 10 − 19 J = 6.206 eV ≅ 6 .21 eV For 300 nm, substitute 6.62 × 10 − 34 m 2 ⋅ kg s for h , 3 × 10 8 m s for c and 300 × 10 − 9 m for λ in equation (1), E photon = ( 6.62 × 10 − 34 m 2 ⋅ kg s ) × ( 3 × 10 8 m s ) 300 × 10 − 9 m = ( 1 b. To determine Wavelength’s of the emission spectrum.

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Question

Chapter 29, Problem 57GP

To determine

To draw: Atom’s energy level diagram.

To determine

Wavelength’s of the emission spectrum.

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Chapter 29, Problem 56GP

Chapter 29, Problem 58GP

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Consider an electron in a hydrogen atom that undergoes a transition from n = 3 to n = 7. a. Is the atom undergoing absorption or emission? b. What is the change in energy of the atom? c. What is the wavelength (in nm) of the photon absorbed or emitted? d. What is the frequency of the photon absorbed or emitted?

The allowed energies of a simple atom are 0.0 eV, 4.0 eV, and 6.0 eV.a. Draw the atom’s energy-level diagram. Label each level with the energy and the principal quantum number.b. What wavelengths appear in the atom’s emission spectrum?c. What wavelengths appear in the atom’s absorption spectrum?

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