You can find the earlier questions (with solution) on Bohr atom model posted on this site by clicking on the label ‘Bohr model’ or ‘hydrogen atom’ below this post. You can get them also by using the search option at the top of this page. Today I give you two questions which are meant for gauging the depth of your understanding of Bohr’s theory.

**(1) When the electron in a hydrogen atom of mass M undergoes transition from an orbit of higher quantum number n_{2} to an orbit of lower quantum number n_{1}, the recoil velocity acquired by the atom is (Rydberg’s constant = R, Planck’s constant = h)**

**(a) ( R/hM) (1/n_{1}^{2} **

**– 1/**

*n*_{2}^{2})

**(b) ( hR/M) (n_{2} **

**–**

*n*_{1})

** (c) 1/ hRM (1/n_{1}^{2} **

**– 1/**

*n*_{2}^{2})

**(d) h/ RM) (1/n_{1}^{2} **

**– 1/**

*n*_{2}^{2})

** (e) ( hR/M) (1/n_{1}^{2} **

**– 1/**

*n*_{2}^{2})

The wave number of the photon emitted because of the electron transition is

ν' = 1/*λ* = *R*(1/*n*_{1}^{2} – 1/*n*_{2}^{2}) where *λ* is the wave length of the photon and *R* is Rydberg’s constant.

The momentum of the photon is *p* = *h**/**λ* = *hR*(1/*n*_{1}^{2} – 1/*n*_{2}^{2}) where *h* is Planck’s constant.

When the photon is emitted with this momentum, the atom recoils (like a gun firing a bullet) with an *equal and opposite* momentum. Therefore, the recoil velocity of the atom is given by

v = *p/M = *(*hR*/*M*)(1/*n*_{1}^{2} – 1/*n*_{2}^{2}).

**(2) If the radius of the innermost electron orbit in a hydrogen atom is R_{1}, the de Broglie wave length of the electron in the second excited state is **

**(a) ****π***R*_{1}** **

**(b) 3****π***R*_{1}** **

**(c) 4****π***R*_{1}** **

**(d) 6****π***R*_{1}** **

**(e) 9****π***R*_{1}** **

The wave length of the electron in the n^{th }orbit is given by

*λ* = 2π*R*_{n}/n where *R*_{n} is the radius of the n^{th} orbit.4

[This follows because the angular momentum of the electron in the n^{th} orbit is

*m*v*R*_{n} = nh/2π.

Therefore, de Broglie wave length, *λ = h*/*m*v = 2π*R*_{n}/n ]

The second excited state has quantum number n = 3 (Third orbit). The radius of the 3^{rd} orbit in terms of the radius *R*_{1} of the first orbit is given by (remembering *R*_{n} = n^{2 }*R*_{1})

*R*_{3} = 9*R _{1}*

Therefore, *λ *= 2π*R*_{n}/n = 2π×9*R*_{1}/3 = 6π*R*_{1}

[It will be convenient to remember that the de Broglie wave length of the electron in the n^{th} orbit is *n times* the the wave length in the innermost orbit].

You will find some useful posts on Atomic Physics and Quantum effects at apphysicsresources

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