(1) B < Al
(2) In < T l
(3) Al < Ga
(4) Ga < In
Atomic Radius Trends in the Periodic Table
The atomic radius generally changes based on the position of an element in the periodic table:
Down a Group: The Atomic radius generally increases due to the addition of electron shells.
Across a Period (Left to Right): The Atomic radius generally decreases due to increasing nuclear charge.
Evaluation of Given Pairs:
B < Al: Boron (B) and Aluminum (Al) are in Group 13. Al is below B, so Al has a larger atomic radius. Correct.
In < Tl: Indium (In) and Thallium (Tl) are in Group 13. Tl is below In, so Tl has a larger atomic radius. Correct.
Al < Ga: Aluminum (Al) and Gallium (Ga) are in Group 13. Ga is below Al, so Ga has a larger atomic radius. Correct.
Ga < In: Gallium (Ga) and Indium (In) are in Group 13. In is below Ga, so In has a larger atomic radius. Correct.
Conclusion: Based on the general trends of atomic radii in the periodic table, all the evaluated orders are correct. The initial reasoning is flawed. Indium, being below Gallium in the same group, has a larger atomic radius.
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to: