Given below are the quantum numbers for 4 electrons.
A. n=3, l=2, ml=1,ms=+\(\frac{1}{2}\)
B. n=4, l=1, ml=0,ms=+\(\frac{1}{2}\)
C. n=4, l=2, ml=–2,ms=–\(\frac{1}{2}\)
D. n=3, l=1, ml=–1,ms=+\(\frac{1}{2}\)
The correct order of increasing energy is
In the circuit shown, assuming the threshold voltage of the diode is negligibly small, then the voltage \( V_{AB} \) is correctly represented by:
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: