The energy of an electron in an orbit of the Bohr's atom is -0.04E_0 text eV$-0.04E_{0} \text{ eV}$ where E_0$E_{0}$ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then frac2pi Lh$\frac{2\pi L}{h}$ is :

### Related Formula E_n = fracE_0n^2$$E_n = \frac{E_0}{n^2}$$ L = fracnh2pi$$L = \frac{nh}{2\pi}$$ ### Core Logic From Bohr's theory, the energy of an electron in the n$n$-th orbit is inversely proportional to n^2$n^2$: E = -fracE_0n^2$$E = -\frac{E_0}{n^2}$$ Equating this to the given energy: -fracE_0n^2 = -0.04 E_0$$-\frac{E_0}{n^2} = -0.04 E_0$$ frac1n^2 = 0.04 = frac125$$\frac{1}{n^2} = 0.04 = \frac{1}{25}$$ n^2 = 25 implies n = 5$$n^2 = 25 \implies n = 5$$ ### Step 1: Final Conclusion Bohr's quantization condition for angular momentum: L = fracnh2pi$$L = \frac{nh}{2\pi}$$ Rearranging for the required term: frac2pi Lh = n$$\frac{2\pi L}{h} = n$$ Since n=5$n=5$, the value is 5$5$. ### Pattern Recognition The expression 2pi L / h$2\pi L / h$ is a direct request for the principal quantum number n$n$. E_n = E_1 / n^2$E_n = E_1 / n^2$ allows finding n$n$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms