પરમાણ્વીય હાઈડ્રોજનના વર્ણપટ રેખાઓની કોઈ આપેલી શ્રેણી માટે \(\Delta \bar V\, = \,\Delta {\bar V_{\max }}\, - \,\Delta {\bar V_{\min }}\) એ \(cm^{-1}\) માં મહતમ અને ન્યૂનતમ આવૃતિઓનો તફાવત છે . \(\Delta {\bar V_{Lymann}}/\,\Delta {\bar V_{Balmer}}\) ગુણોતર જણાવો.

A \(5:4\)
B \(4:1\)
C \(9:4\)
D \(27:5\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(9:4\)

Step-by-step Solution

Detailed explanation

\(\Delta {{\bar V}_{Lymann}} = \frac{{{R_H}}}{4}\) \(\Delta {{\bar V}_{Balmer}} = \frac{{{R_H}}}{9}\) \( \Rightarrow \frac{{\Delta {{\bar V}_{Lymann}}}}{{\Delta {{\bar V}_{Balmer}}}} = \frac{9}{4}\) Formula…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

સૂચિ-\(I\) ને સૂચિ-\(II\) સાથે જોડો.

સૂચિ - I (પ્રક્રિયા)	સૂચિ - II (રેડોક્સ પ્રક્રિયાનો પ્રકાર)
(A) \(N _{2(g)}+ O _{2(g)} \rightarrow 2 NO _{( g )}\)	(I) વિઘટન
(B) \(2 Pb\left( NO _3\right)_{2(s)} \rightarrow 2 PbO _{( s )}+4 NO _{2(g)}+ O _{2(g)}\)	(II) વિસ્થાપન
(C) \(2 Na _{( s )}+2 H _2 O _{( l )} \rightarrow 2 NaOH _{( aq .)}+ H _{2(g)}\)	(III) અસમાનુપાતન
(D) \(2 NO _{2(g)}+2- OH _{( aq )} \rightarrow NO _{2( aq .)}^{-}+ NO _{3( qq .)}^{-}+ H _2 O _{( l )}\)	(IV) સંયોગીકરણ

નીચે આપેલા વિકલ્પોમાંથી સાચો જવાબ પસંદ કરો:

JEE Mains 2024 Medium

અનનિલનિયમનો પરમાણ્વીય ક્રમાંક ...........છે.

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	\(1\)	\(2\)	\(3\)
\(\text{HI} \left(\text{molL} ^{-1}\right)\)	\(0.005\)	\(0.01\)	\(0.02\)
Rate \(\left(\text{molL} ^{-1}\text s-1\right)\)	\(7.5 \times 10^{-4}\)	\(3.0 \times 10^{-3}\)	\(1.2 \times 10^{-2}\)