- Statement I: This is a restatement of the Heisenberg Uncertainty Principle, which asserts that it is impossible to precisely determine both the position and the momentum of a particle simultaneously. This statement is true.
- Statement II: The Heisenberg Uncertainty Principle provides the relationship between the uncertainty in position (\( \Delta x \)) and momentum (\( \Delta p \)), given by: \[ \Delta x \Delta p \geq \frac{h}{4\pi} \] For an electron, if the uncertainties in position and momentum are equal, the uncertainty in velocity \( \Delta v \) can be expressed as: \[ \Delta v = \frac{\Delta p}{m} \geq \sqrt{\frac{h}{\pi}} \times \frac{1}{2m} \] This statement is also correct.
Therefore, the correct answer is \( \boxed{(2)} \) Both Statement I and Statement II are true.
Match the LIST-I with LIST-II
LIST-I (Energy of a particle in a box of length L) | LIST-II (Degeneracy of the states) | ||
---|---|---|---|
A. | \( \frac{14h^2}{8mL^2} \) | I. | 1 |
B. | \( \frac{11h^2}{8mL^2} \) | II. | 3 |
C. | \( \frac{3h^2}{8mL^2} \) | III. | 6 |
Choose the correct answer from the options given below:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.