JEE Advanced · Chemistry · 18. Chemical Kinetics
In the decay sequence:
and are the particles/ radiation emitted by the respective isotopes. The correct option(s) is/are:
- A is an isotope of uranium.
- B is .
- C will deflect towards the negatively charged plate.
- D is -ray.
Answer & Solution
Correct Answer
(A) is an isotope of uranium.
Step-by-step Solution
Detailed explanation
\({ }_{92}\text U \xrightarrow{-\frac{4}{2} \propto}{ }_{90}^{23} \text{Th} \xrightarrow{^{-0}_{-1} \beta}{ }_{91}^{234} \text{Pa} \xrightarrow{^{-0}_{-1} \beta}{ }_{92}^{234}\)\(\text U \xrightarrow{-\frac{4}{2} \propto}{ }_{90}^{234} \text{Th}\)
\(\text{I.e., }\text{x}_1=\alpha,\text x_2=\beta,\text x_3=\beta\) and \(\text x_4=\alpha\).
\(\text{I.e., }\text{x}_1=\alpha,\text x_2=\beta,\text x_3=\beta\) and \(\text x_4=\alpha\).
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Chemistry
- Match the compounds in LIST- with the observations in LIST-, and choose the correct option.
LIST- LIST- Aniline Sodium fusion extract of the compound on boiling with , followed by acidification with conc. , gives Prussian blue color. Cresol Sodium fusion extract of the compound on treatment with sodium nitroprusside gives blood red color. Cysteine Addition of the compound to a saturated solution of results in effervescence. Caprolactam The compound reacts with bromine water to give a white precipitate. Treating the compound with neutral solution produces violet color. JEE Advanced 2022 Hard - Which of the following halides react(s) with \(\mathrm{AgNO}_{3}(\mathrm{aq})\) to give a precipitate that dissolves in \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq})\) ?JEE Advanced 2012 Medium
- Consider a reaction \(a G+b H \rightarrow\) products. When concentration of both the reactants \(G\) and \(H\) is doubled, the rate increases by eight times. However, when concentration of \(G\) is doubled keeping the concentration of \(H\) fixed, the rate is doubled. The overall order of the reaction isJEE Advanced 2007 Easy
- Among the complex ions,
\(\left[\mathrm{Co}\left(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\right)_{2} \mathrm{Cl}_{2}\right]^{+},\)\(\left[\mathrm{CrCl}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]^{3-},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}(\mathrm{OH})_{2}\right]^{+}\), \(\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{CN})_{4}\right]^{-},\)\(\left[\mathrm{Co}(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2})_{2}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}\right]^{2+}\) and \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\), the number of complex ion(s) that show(s) cis-trans isomerism isJEE Advanced 2015 Medium - The oxidation number of \(\mathrm{Mn}\) in the product of alkaline oxidative fusion of \(\mathrm{MnO}_2\) isJEE Advanced 2009 Medium
- For the reaction:
The correct statement(s) in the balanced equation is/are:JEE Advanced 2014 Medium
More PYQs from JEE Advanced
- For the function \(f(x)=x \cos \frac{1}{x}, x \geq 1\),JEE Advanced 2009 Hard
- Match each of the diatomic molecules in Column I with its property/properties in Column II.
Column I Column II A. \(B _2\) p. Paramagnetic B. \(N_2\) q. Undergoes oxidation C. \(O _2^{-}\) r. Undergoes reduction D. \(O _2\) s. Bond order t. Mixing of 's' and 'p' orbitals JEE Advanced 2009 Medium - Four solid spheres each of diameter \(\sqrt{5} \mathrm{~cm}\) and mass \(0.5 \mathrm{~kg}\) ar placed with their centres at the corners of a square of side \(4 \mathrm{~cm}\). The moment of inertia of the system about the diagonal of the square is \(N \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2\), then \(N\) isJEE Advanced 2011 Medium
- Consider the lines \(L_{1}\) and \(L_{2}\) defined by
\[
L_{1}: x \sqrt{2}+y-1=0 \text { and } L_{2}: x \sqrt{2}-y+1=0
\]
For a fixed constant \(\lambda\), let \(C\) be the locus of a point \(P\) such that the product of the distance of \(P\) from \(L_{1}\) and the distance of \(P\) from \(L_{2}\) is \(\lambda^{2}\). The line \(y=2 x+1\) meets \(C\) at two points \(R\) and \(S\), where the distance between \(R\) and \(S\) is \(\sqrt{270}\).
Let the perpendicular bisector of \(R S\) meet \(C\) at two distinct points \(R^{\prime}\) and \(S^{\prime}\). Let \(D\) be the square of the distance between \(R^{\prime}\) and \(S^{\prime}\).
The value of isJEE Advanced 2021 Hard - A long circular tube of length \(10 \mathrm{~m}\) and radius \(0.3 \mathrm{~m}\) carries a current I along its curved surface as shown. A wire loop of resistance \(0.005 \Omega\) and of radius \(0.1 \mathrm{~m}\) is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as \(I=I_0 \cos 300 t\), where \(I_0\) is constant. If the magnetic moment of the loop is \(N \mu_0 I_0 \sin 300 t\), then \(N\) is
JEE Advanced 2011 Hard - The carbon-based reduction method is NOT used for the extraction ofJEE Advanced 2013 Easy