Questions on Bohr model of hydrogen-like atoms usually appear in medical and engineering and other degree entrance examinations. You can access all posts on Bohr model on this site by clicking on the label ‘Bohr model’ below this post. Here are a few more questions on Bohr model of hydrogen atom:

**(1)** The electron in a hydrogen atom in the ground state absorbs energy equal to 12.1 eV and gets elevated to the highest possible excited state. What will be change in the angular momentum of the electron? (h = Planck’s constant)

(a) h/π

(b) 2h/π

(c) 3h/π

(d) 4h/π

(e)** **h/2π

In the case of the hydrogen atom the energies of the electron are – 13.6 eV, – 3.4 eV, – 1.51 eV, – 0.85 eV etc. for values of *n* equal to 1, 2, 3, 4 etc. respectively. After absorbing 12.1 eV of energy, the electron in the innermost orbit having energy – 13.6 eV gets elevated to the 3^{rd} orbit where its energy is – 1.51 eV.

The angular momentum of the electron in the innermost orbit (of quantum no. n = 1) is h/2π. In the 3^{rd} orbit its angular momentum is 3h/2π. Therefore, the change in the angular momentum is (3h/2π) – (h/2π) = h/π

**(2) **If the ionisation potential of hydrogen atom is 13.6 volt, the energy required to remove an electron from the second orbit of hydrogen atom is

(a) 0.54 eV

(b) 0.85 eV

(c) 1.51 eV

(d) 3.4 eV

(e)** **13.6 eV

By stating that the ionisation potential of hydrogen atom is 13.6 volt, you are informed that 13.6 eV of energy is required to remove an electron from the innermost orbit of the hydrogen atom. This is because of the energy –13.6 eV possessed by the electron in the innermost orbit. Since the electron in the 2^{nd} orbit possesses energy equal to –13.6/2^{2} = –3.4 eV, the energy required to remove an electron from the second orbit of hydrogen atom is 3.4 eV.

**(3) **In the hydrogen atom the transition that gives radiation in the visible region is

(a) from n > 1 to n = 1

(b) from n > 1 to n = 1

(c) from n > 2 to n = 1

(d) from n > 3 to n = 1

(e)** **from n > 2 to n = 2** **** **

The hydrogen atom gives visible spectral lines in the balmer series because of the transitions from outer orbits to the 2^{nd} orbit. So the correct option is (e).

**(4) **In the hydrogen spectrum the frequency of a line resulting from the transition of the electron from the orbit of quantum number *n*_{x }to quantum number *n*_{1} is *f*. In a hydrogen- like atom the *same transition* gives rise to a spectral line of frequency 9*f*. The hydrogen- like atom has atomic number

(a) 1

(b) 2

(c) 3

(d) 6

(e)** **9** **

The energy (*E*_{n})_{ }of the electron in the orbit of quantum number *n* in a hydrogen-like atom is given by

*E*_{n} = –13.6 *z*^{2}/n^{2} where *z* is the atomic number.

The energy difference between two states of the hydrogen-like atom is therefore *z*^{2} times the energy difference in the case of the hydrogen atom. The frequency of the resulting spectral line also is *z*^{2} times the frequency in the case of the hydrogen atom. Since the frequency is 9 times, the atomic number of the hydrogen like atom is 3.

[The hydrogen-like atom in this question is doubly ionised lithium (Li^{++})].

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