Class 12th

Every element has a unique set of spectral lines as its electrons occupy specific energy levels. What scientists know for certain is that these unique patterns act like fingerprints. It helps in identifying elements in stars, flames, or unknown samples. For instance, Helium was discovered in the Sun's spectrum before it was found on Earth.

We see a discrete emission spectrum when electrons inside excited atoms or ions in gases fall back from higher energy levels to lower ones. Every transition releases a photon of a specific wavelength. Usually the spectrum appears as sharp, bright lines. Dark gaps separate them. These lines are unique to each element.

On the other hand, a continuous emission spectrum is produced when hot solids, liquids, or dense gases emit radiation. This happens because of the collective motion of their atoms and electrons. We don't see sharp lines. Instead the spectrum shows a smooth and unbroken spread of all wavelengths. 

The truth table for the logical expression (p ∧ q) → (p → q) is as follows:

The final column shows that the expression is a tautology, meaning it is always true regardless of the truth values of p and q.

Please consider the following Image

o   Let the number of atoms of element A be N.

o   Tetrahedral voids = 2N.

o   Atoms of element M = (2/3) * 2N = 4/3 N.

o   The ratio M : A is (4/3 N) : N, which simplifies to 4:3.

o   So, the formula of the compound is M? A?

The equation of a plane parallel to x - 2y + 2z - 3 = 0 is x - 2y + 2z + λ = 0.
The distance from the point (1, 2, 3) to this plane is 1.
|1 - 2 (2) + 2 (3) + λ| / √ (1² + (-2)² + 2²) = 1
|1 - 4 + 6 + λ| / √9 = 1
|3 + λ| / 3 = 1
|3 + λ| = 3
3 + λ = 3 or 3 + λ = -3
λ = 0 or λ = -6.

1 = (2-1)¹ (The n is likely 1).
3? = (7-4)³ (This seems to be a pattern matching (a-b)^c).
4²? = (12-8)? ! = 4²?
The blank space must be (5-3)² = 2² = 4.

Circle 1: x² + y² - 10x - 10y + 41 = 0
Center C? = (5,5).
Radius r? = √ (5² + 5² - 41) = √ (25+25-41) = √9 = 3.

Circle 2: x² + y² - 22x - 10y + 137 = 0
Center C? = (11,5).
Radius r? = √ (11² + 5² - 137) = √ (121+25-137) = √9 = 3.

Distance between centers d (C? , C? ) = √ (11-5)² + (5-5)²) = √ (6²) = 6.

Sum of radii r? + r? = 3 + 3 = 6.
Since the distance between the centers is equal to the sum of their radii, the circles touch externally at one point.

Linear: N? (azide ion)

Bent-shape: O? (ozone), NO? (nitrite ion), Cl? O (dichlorine monoxide)