Explanation If an electron jumps from n = 3 shell to the n = 1 shell, then the resulting photon is a K β X-ray. The expression for the energy of a K-shell electron is, E K = − 13.6 eV ( Z − 1 ) 2 1 2 Here, E K is the energy of K-shell electron. Z is the atomic number of molybdenum. Substitute 42 for Z in the above expression to get E K . E K = − 13.6 eV ( Z − 1 ) 2 1 2 = − 22861.6 eV The expression for the energy of an M-shell electron is, E M = − 13.6 eV ( Z − 9 ) 2 3 2 Here, E M is the energy of M-shell electron. Substitute 42 for Z in the above expression to get E M . E M = − 13.6 eV ( Z − 9 ) 2 3 2 = − 1645.6 eV The formula to calculate the change in energy of an electron that jumps from the M-shell to K-shell is, | Δ E | = | E K − E M | Here, Δ E is the change in energy. Substitute − 22861.6 eV for E K and − 1645.6 eV for E M in the above expression to get Δ E . | Δ E | = | E K − E M | = | − 22861

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The Bohr model

Students often get confused between Rutherford’s model and Bohr model of an atom. They interpret the postulates of both these models to be the same, which is incorrect. Rutherford’s model differs from Bohr’s model in the following ways:

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Chapter 31, Problem 78GP

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The approximate wavelength of Kβ X-rays emitted by molybdenum and comparison of the calculated wavelength with Figure 31-27.

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Chapter 31, Problem 77GP

Chapter 31, Problem 79GP

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