You will usually find a couple of questions pertaining to Bohr model of hydrogen atom in most entrance test papers. Questions in this section are interesting and often simple. Consider the following M.C.Q.:

Now consider the following questions which appeared in the Kerala Engineering Entrance Test of 2002:

Since the angular momentum is nh/2π, on substituting for h (=6.63×10

**Which state of the triply ionised Be+++ has the same orbital radius (for the electron) as that of the hydrogen atom in the ground state?**

(a) First excited state (b) Second excited state (c) Third excited state (d) Fourth excited state (e) Ground state

The Be atom has lost 3 electrons since it is triply ionised and it behaves as a hydrogen like atom. Therefore, the radius of the orbit of quantum number ‘n’ is r n(a) First excited state (b) Second excited state (c) Third excited state (d) Fourth excited state (e) Ground state

^{2}/z where ‘r’ is the Bohr radius (radius of the electron orbit in the hydrogen atom in the ground state) and ‘z’ is the atomic number of the hydrogen like atom. Since z = 4 for Be, we have r = r n^{2}/4 from which n = 2, which means the first excited state. So, the correct option is (a).Now consider the following questions which appeared in the Kerala Engineering Entrance Test of 2002:

**(1) The energy of an electron in excited hydrogen atom is -3.4 eV. Then, according to Bohr’s theory, the angular momentum of the electron in Js is**

(a) 2.11×10

Since the energy of the electron is -3.4 eV, the electron is in the second orbit. This is checked using the energy (in eV) expression E = -13.6/n(a) 2.11×10

^{-34}(b) 3×10^{-34}(c) 3×10^{-34}(d) 0.5×10^{-34}^{2}= -3.4 eV, when n = 2.Since the angular momentum is nh/2π, on substituting for h (=6.63×10

^{-34}Js) and n (=2) we obtain the first option as the correct answer.**(2) Consider the spectral line resulting from the transition from n = 2 to n = 1, in atoms and ions given below. The shortest wave length is produced by**

(a) hydrogen atom (b) deuterium atom (c) singly ionized helium (d) doubly ionized helium (e) doubly ionized lithium(a) hydrogen atom (b) deuterium atom (c) singly ionized helium (d) doubly ionized helium (e) doubly ionized lithium

**The correct option is (e). The expression for the energy of an electron in a hydrogen like atom is E = (- 13.6 z**

^{2})/n

^{2}. The energy for a given value of ‘n’ is maximum for lithium in the present case (since z = 3). The energy difference also is therefore maximum (for the given transition) in the case of lithium, giving rise to the shortest wavelength.

You will find more questions with solution at AP Physics Resources: AP Physics B– Multiple Choice Questions (for practice) on Atomic Physics and Quantum Effects

Hi there!

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With best wishes

-MV(physicsplus.blogspot.com)