Atoms & Nuclei

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Question 32 of 239

The energy of an electron in an orbit of the Bohr's atom is −0.04E0eV-0.04 E_{0} \mathrm{eV} where E0E_{0} is the ground state energy. If LL is the angular momentum of the electron in this orbit and hh is the Planck's constant, then 2πLh\frac{2 \pi L}{h} is ____\_\_\_\_ :

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Answer: (A) 5

In Bohr's atom, the energy levels are En=−E0n2E_n = -\frac{E_0}{n^2} where E0E_0 is the ground state energy.
Given E=−0.04E0E = -0.04E_0, we have −E0n2=−0.04E0-\frac{E_0}{n^2} = -0.04E_0, which gives n2=25n^2 = 25, so n=5n = 5.
The angular momentum in Bohr's model is L=nℏ=nh2πL = n\hbar = n\frac{h}{2\pi}.
Therefore, 2πLh=2πh×nh2π=n=5\frac{2\pi L}{h} = \frac{2\pi}{h} \times n\frac{h}{2\pi} = n = 5.

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