Question 5

Option (A) is correct: Heisenberg's Uncertainty Principle states that $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$, which prevents the determination of a precise trajectory or definite path for an electron. Option (B) is correct: The energy of an electron at infinite distance from the nucleus is taken as zero. Since the energy of a bound electron is negative ($E_n = -13.6 \frac{Z^2}{n^2}$ eV), any orbital like $2s$ has energy lower than zero. Option (C) is correct: In Bohr's model, the energy $E_n$ is inversely proportional to $n^2$ with a negative sign. Thus, $n=1$ yields the most negative value, which represents the lowest energy and most stable state (ground state). Option (D) is incorrect: According to Bohr's model, the velocity of an electron $v_n$ is given by $v_n = \frac{2\pi kZe^2}{nh}$, which means $v_n \propto \frac{1}{n}$. Therefore, velocity decreases as $n$ increases.