The energy of an electron in an orbit of the Bohr’s atom is -0.04E₀ eV where E₀ 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 2πL/h i $s __________$ :
Options

Question

The energy of an electron in an orbit of the Bohr’s atom is -0.04E₀ eV where E₀ 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 2πL/h i $s __________$ :

Options

G1 Physics · Accepted Answer

For a hydrogen-like Bohr orbit, the magnitude of the energy varies as |Eₙ| = |E₁|/n². The given orbital energy has magnitude 0.04 times the ground-state magnitude. Therefore, 1/n² = 0.04 = 1/25. Hence, n² = 25, so n = 5. Bohr's angular-momentum condition is L = nh/(2π). Therefore, 2πL/h = n = 5. Correct option: C.

The energy of an electron in an orbit of the Bohr’s atom is -0.04E₀ eV where E₀ 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 2πL/h i s __________ :