The maximum number of bright fringes depends on the wavelength and the slit separation, as it determines how many integer multiples of the wavelength fit within the observable range.

The d subshell can hold a maximum of 10 electrons.

The f-subshell can hold a maximum of 14 electrons.

The d subshell can hold a maximum of 10 electrons.

The p subshell can hold a maximum of 6 electrons.

The maximum number of electrons in an energy level is given by the formula 2n². For n=3, it is 2(3)² = 18.

The maximum number of electrons in a shell is given by the formula 2n². For n=3, it is 2(3)² = 18.

The maximum number of electrons in a shell is given by the formula 2n². For n=3, it is 2(3)² = 18.

The maximum number of electrons in an energy level is given by 2n^2. For n=4, it is 2(4^2) = 32.

Phosphorus can exhibit a maximum oxidation state of +5 in compounds like PCl5.

The maximum profit occurs at x = -b/(2a) = 10/2 = 5. P(5) = -5^2 + 10*5 - 16 = 9.

The vertex occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 5.

The vertex occurs at x = 3. f(3) = -9 + 18 - 8 = 1.

Maximum wavelength (λ) = hc/φ. λ = (1240 eV.nm) / 4.0 eV = 310 nm.

Using the distance formula from a point to a line, the minimum distance is calculated to be 2 units.

The minimum energy required to remove an electron is equal to the work function of the metal, which is 2.5 eV.

The minimum energy required is equal to the work function, which is 2 eV.

The minimum energy required to remove an electron is equal to the work function of the metal, which is 2.5 eV.

For the first order of bright fringe in reflected light, the minimum thickness of the soap bubble is λ/4 due to the phase change upon reflection.

For constructive interference in a soap bubble, the minimum thickness should be λ/4, accounting for the phase change upon reflection.

The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 12 = 0.

The vertex occurs at x = 2. f(2) = -5.

The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 9 = 0.

The vertex form gives the minimum at x = 2. f(2) = 2.

The vertex occurs at x = 2. f(2) = 2.

The vertex form gives the minimum at x = 2. f(2) = 2, thus minimum value is 3.

The vertex form gives the minimum at x = 3. f(3) = 3^2 - 6(3) + 10 = 1.

The vertex gives the minimum at x = 2. f(2) = 4(2^2) - 16(2) + 20 = 4.

The vertex occurs at x = 2. f(2) = 4(2^2) - 16(2) + 20 = 4.

The modulus is calculated as √(3^2 + 4^2) = √(9 + 16) = √25 = 5.