Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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

Last updated date: 13th Jun 2024
Total views: 373.8k
Views today: 10.73k
Verified
373.8k+ views
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, $E=-\dfrac{13.6}{{{n}^{2}}}$

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).