Question 31

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

Energy of electron in orbit: $$E_n = \frac{E_0}{n^2}$$ (where $$E_0$$ is ground state energy, negative).

Given: $$E_n = -0.04 E_0$$, where $$E_0$$ is the magnitude of ground state energy.

So $$\frac{E_0}{n^2} = 0.04 E_0$$ (in terms of magnitudes).

$$n^2 = \frac{1}{0.04} = 25$$

$$n = 5$$

Angular momentum: $$L = \frac{nh}{2\pi}$$

$$\frac{2\pi L}{h} = n = 5$$

The answer is Option 3: 5.

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