Explanation Given info: The values of n for neutral hydrogen, singly ionized helium and doubly ionized lithium are 3 , 2 and 1 respectively. (A) Formula to calculate the Bohr energy is, E n = − 13.6 Z 2 n 2 eV Here, E n is the energy of n t h orbit. Z is the atomic number of an atom. n is the number of orbit. The value of Z for hydrogen is 1 . Substitute the values of Z and n in the above expression to get E n . E 3 = − 13.6 ( 1 ) 2 ( 3 ) 2 eV = − 13.6 9 eV = − 1 .51 eV Thus, the energy of third orbit in hydrogen atom is − 1 .51 eV . Similarly, (B) Calculate the Bohr energy for singly ionized helium atom. The value of Z for helium is 2 . Substitute the values of Z and n in the expression of Bohr energy. E 2 = − 13.6 ( 2 ) 2 ( 2 ) 2 eV = − 54

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Chapter 31, Problem 13PCE

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The order of increasing energy for given substances.

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Chapter 31, Problem 12PCE

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Do the various spectral lines of the hydrogen atom overlap?
A 4.653-urn emission line of atomic hydrogen corresponds to transition between the states nf=5 and ni. Find ni.
Consider the Bohr model as applied to the following threeatoms: (A) neutral hydrogen in the state n = 2; (B) singly ionizedhelium in the state n = 1; (C) doubly ionized lithium in the staten = 3. Rank these three atoms in order of increasing Bohr radius.Indicate ties where appropriate.

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