उत्कृष्ट गैसों की आयनन ऊर्जा सर्वोच्च क्यो होती है समझाइए |
Answers
Answer:
We have given that,
\displaystyle{\sf\:3\:\cot\:A\:=\:4}3cotA=4
We have to find the value of
\displaystyle{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}}
3cosA−2sinA
3cosA+2sinA
Now,
\displaystyle{\sf\:3\:\cot\:A\:=\:4}3cotA=4
\displaystyle{\implies\boxed{\sf\:\cot\:A\:=\:\dfrac{4}{3}}}⟹
cotA=
3
4
Consider a right triangle ABC right angled at angle B.
We know that,
\displaystyle{\pink{\sf\:\cot\:A\:=\:\dfrac{Adjacent\:side}{Opposite\:side}}}cotA=
Oppositeside
Adjacentside
\displaystyle{\implies\sf\:\dfrac{4}{3}\:=\:\dfrac{AB}{BC}}⟹
3
4
=
BC
AB
\displaystyle{\implies\sf\:AB\:=\:4x}⟹AB=4x
\displaystyle{\sf\:BC\:=\:3x}BC=3x
Now, by Pythagoras theorem,
\displaystyle{\pink{\sf\:(\:AC\:)^2\:=\:(\:AB\:)^2\:+\:(\:BC\:)^2}}(AC)
2
=(AB)
2
+(BC)
2
\displaystyle{\implies\sf\:AC^2\:=\:(\:4x\:)^2\:+\:(\:3x\:)^2}⟹AC
2
=(4x)
2
+(3x)
2
\displaystyle{\implies\sf\:AC^2\:=\:16x^2\:+\:9x^2}⟹AC
2
=16x
2
+9x
2
\displaystyle{\implies\sf\:AC^2\:=\:25x^2}⟹AC
2
=25x
2
\displaystyle{\implies\sf\:AC\:=\:5x\:}⟹AC=5x
\displaystyle{\therefore\boxed{\sf\:Hypotenuse\:=\:5x}}∴
Hypotenuse=5x
Now, we know that,
\displaystyle{\pink{\sf\:\sin\:A\:=\:\dfrac{Opposite\:side}{Hypotenuse}}}sinA=
Hypotenuse
Oppositeside
\displaystyle{\implies\sf\:\sin\:A\:=\:\dfrac{3\:\cancel{x}}{5\:\cancel{x}}}⟹sinA=
5
x
3
x
\displaystyle{\implies\sf\:\sin\:A\:=\:\dfrac{3}{5}}⟹sinA=
5
3
\displaystyle{\implies\sf\:2\:\sin\:A\:=\:\dfrac{2\:\times\:3}{5}}⟹2sinA=
5
2×3
\displaystyle{\implies\boxed{\blue{\sf\:2\:\sin\:A\:=\:\dfrac{6}{5}}}}⟹
2sinA=
5
6
Now, we know that,
\displaystyle{\pink{\sf\:\cos\:A\:=\:\dfrac{Adjacent\:side}{Hypotenuse}}}cosA=
Hypotenuse
Adjacentside
\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{4\:\cancel{x}}{5\:\cancel{x}}}⟹cosA=
5
x
4
x
\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{4}{5}}⟹cosA=
5
4
\displaystyle{\implies\sf\:3\:\cos\:A\:=\:\dfrac{3\:\times\:4}{5}}⟹3cosA=
5
3×4
\displaystyle{\implies\boxed{\green{\sf\:3\:\cos\:A\:=\:\dfrac{12}{5}}}}⟹
3cosA=
5
12
Now, we have to find the value of
\displaystyle{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}}
3cosA−2sinA
3cosA+2sinA
\displaystyle{\implies\sf\:\dfrac{\dfrac{12}{5}\:+\:\dfrac{6}{5}}{\dfrac{12}{5}\:-\:\dfrac{6}{5}}}⟹
5
12
−
5
6
5
12
+
5
6
\displaystyle{\implies\sf\:\dfrac{\dfrac{12\:+\:6}{5}}{\dfrac{12\:-\:6}{5}}}⟹
5
12−6
5
12+6
\displaystyle{\implies\sf\:\dfrac{\dfrac{18}{5}}{\dfrac{6}{5}}}⟹
5
6
5
18
\displaystyle{\implies\sf\:\dfrac{18}{\cancel{5}}\:\times\:\dfrac{\cancel{5}}{6}}⟹
5
18
×
6
5
\displaystyle{\implies\sf\:\cancel{\dfrac{18}{6}}}⟹
6
18
\displaystyle{\implies\sf\:3}⟹3
\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}\:=\:3\:}}}}∴
3cosA−2sinA
3cosA+2sinA
=3
Explanation:
We have given that,
\displaystyle{\sf\:3\:\cot\:A\:=\:4}3cotA=4
We have to find the value of
\displaystyle{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}}
3cosA−2sinA
3cosA+2sinA
Now,
\displaystyle{\sf\:3\:\cot\:A\:=\:4}3cotA=4
\displaystyle{\implies\boxed{\sf\:\cot\:A\:=\:\dfrac{4}{3}}}⟹
cotA=
3
4
Consider a right triangle ABC right angled at angle B.
We know that,
\displaystyle{\pink{\sf\:\cot\:A\:=\:\dfrac{Adjacent\:side}{Opposite\:side}}}cotA=
Oppositeside
Adjacentside
\displaystyle{\implies\sf\:\dfrac{4}{3}\:=\:\dfrac{AB}{BC}}⟹
3
4
=
BC
AB
\displaystyle{\implies\sf\:AB\:=\:4x}⟹AB=4x
\displaystyle{\sf\:BC\:=\:3x}BC=3x
Now, by Pythagoras theorem,
\displaystyle{\pink{\sf\:(\:AC\:)^2\:=\:(\:AB\:)^2\:+\:(\:BC\:)^2}}(AC)
2
=(AB)
2
+(BC)
2
\displaystyle{\implies\sf\:AC^2\:=\:(\:4x\:)^2\:+\:(\:3x\:)^2}⟹AC
2
=(4x)
2
+(3x)
2
\displaystyle{\implies\sf\:AC^2\:=\:16x^2\:+\:9x^2}⟹AC
2
=16x
2
+9x
2
\displaystyle{\implies\sf\:AC^2\:=\:25x^2}⟹AC
2
=25x
2
\displaystyle{\implies\sf\:AC\:=\:5x\:}⟹AC=5x
\displaystyle{\therefore\boxed{\sf\:Hypotenuse\:=\:5x}}∴
Hypotenuse=5x
Now, we know that,
\displaystyle{\pink{\sf\:\sin\:A\:=\:\dfrac{Opposite\:side}{Hypotenuse}}}sinA=
Hypotenuse
Oppositeside
\displaystyle{\implies\sf\:\sin\:A\:=\:\dfrac{3\:\cancel{x}}{5\:\cancel{x}}}⟹sinA=
5
x
3
x
\displaystyle{\implies\sf\:\sin\:A\:=\:\dfrac{3}{5}}⟹sinA=
5
3
\displaystyle{\implies\sf\:2\:\sin\:A\:=\:\dfrac{2\:\times\:3}{5}}⟹2sinA=
5
2×3
\displaystyle{\implies\boxed{\blue{\sf\:2\:\sin\:A\:=\:\dfrac{6}{5}}}}⟹
2sinA=
5
6
Now, we know that,
\displaystyle{\pink{\sf\:\cos\:A\:=\:\dfrac{Adjacent\:side}{Hypotenuse}}}cosA=
Hypotenuse
Adjacentside
\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{4\:\cancel{x}}{5\:\cancel{x}}}⟹cosA=
5
x
4
x
\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{4}{5}}⟹cosA=
5
4
\displaystyle{\implies\sf\:3\:\cos\:A\:=\:\dfrac{3\:\times\:4}{5}}⟹3cosA=
5
3×4
\displaystyle{\implies\boxed{\green{\sf\:3\:\cos\:A\:=\:\dfrac{12}{5}}}}⟹
3cosA=
5
12
Now, we have to find the value of
\displaystyle{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}}
3cosA−2sinA
3cosA+2sinA
\displaystyle{\implies\sf\:\dfrac{\dfrac{12}{5}\:+\:\dfrac{6}{5}}{\dfrac{12}{5}\:-\:\dfrac{6}{5}}}⟹
5
12
−
5
6
5
12
Answer:
Explanation:
उत्कृष्ट गैसों की आयनन ऊर्जा सर्वाधिक होती है क्योंकि
(i) प्रत्येक आवर्त में उत्कृष्ट गैसों का इलेक्ट्रॉनिक विन्यास सर्वाधिक स्थायी (ns2 np6) होता है,
(ii) तत्व, जैसे Be (1s2 2s2) तथा मैग्नीशियम (1s2 2s2 2p6, 3s2) में कक्षकें पूर्ण भरी पायी जाती हैं।
(iii) तत्व, जैसे N (1s2 2s2 2px1 2py1 2pz1 ) तथा P (1s2 2s2 2p6 3s2 3px1 3py1 3pz1) में कक्ष के समान उपकोश में अर्द्ध-पूरित हैं, अत: ये स्थायी हैं। अत: इनसे इलेक्ट्रॉनों के निष्कासन के लिए अधिक ऊर्जा की आवश्यकता होगी और ये भी एक कारन है जिनके कारन इनकी आयनन ऊर्जा उच्च होती है |