Question

If there are two de-Broglie waves formed in an orbit of H- atom then the energy of electron in that orbit is: -
(A) -3.4 eV
(B) -13.6 eV
(C) -9.6 eV
(D) 1.51 eV

Answer 1

Hint: Use the Bohr’s postulate for hydrogen atom (given below) to answer this question.
(1) an electron moves around the nucleus in a circular orbit,
(2) an electron’s angular momentum in the orbit is quantized,
 $ L=n\left( \dfrac{h}{2\pi } \right) $
where h is the planck's constant. [h = 6.6 x 10-34 J/s].
(3) the change in an electron’s energy as it makes a quantum jump from one orbit to another is always accompanied by the emission or absorption of a photon.
Use the formula of energy of electron of hydrogen atom in its nth orbit, $ E=-\dfrac{13.6}{{{n}^{2}}} $

Complete step by step answer
Wavelength of de-Broglie wave, $ \lambda =\dfrac{h}{mv} $
circumference of the orbit= 2πr
We have given 2 de-Broglie waves formed in the orbit, therefore
 $ 2\pi r=2.\dfrac{h}{mv} $
Rearranging the above equation, we get;
 $ \begin{align}
  & \Rightarrow mvr=\dfrac{h}{\pi }=2.\left( \dfrac{h}{2\pi } \right) \\
 & \Rightarrow L=2.\left( \dfrac{h}{2\pi } \right) \\
\end{align} $ `
Comparing above equation with the Bohr’s second postulate i.e., $ 2\pi r=2.\dfrac{h}{mv} $ $ L=n\left( \dfrac{h}{2\pi } \right) $
We get, n=2.
energy of electron of hydrogen atom in its nth orbit, $ E=-\dfrac{13.6}{{{n}^{2}}} $ .
Therefore, $ E=\dfrac{-13.6}{{{2}^{2}}}=-3.4\text{ eV} $
Hence, the correct option is (A).

Additional information
The Bohr model of hydrogen was the first model of atomic structure to correctly explain the radiation spectra of atomic hydrogen. It was preceded by the Rutherford nuclear model of the atom.
In Rutherford’s model, an atom consists of a positively charged point-like nucleus that contains almost the entire mass of the atom and of negative electrons that are located far away from the nucleus.

Note
Remember the following points regarding the Bohr’s postulates: -
(1) Bohr’s postulates were only applicable in the case of hydrogen atoms. It couldn’t elaborate spectra of multi-electron atoms
(2) Wave nature of the electron was not justified by the model.
(3) It is not in accordance with the Heisenberg’s uncertainty principle, which said that it is impossible to evaluate the precise position and momentum of microscopic particles. Only their probability could be estimated.