Radioactive Decay Series JEE 2019: B, C and D Explained

What is the radioactive decay series JEE 2019 question?

B, C and D are correct: the missing intermediate is uranium-234, and the third emission is beta-minus, not gamma. The radioactive decay series JEE 2019 question is JEE Advanced 2019, Paper 1, Chemistry, Q.8, from Atomic Structure. It is a multiple-correct MCQ, tagged medium in the question bank, with a suggested solving time of 90 seconds, not an official time limit.

Identify every claim consistent with the particles or radiation emitted at the four arrows. More than one option may be correct:

_{92}^{238} U \overset{- x_{1}}{\to}_{90}^{234} T h \overset{- x_{2}}{\to}_{91}^{234} P a \overset{- x_{3}}{\to}^{234} Z \overset{- x_{4}}{\to}_{90}^{230} T h

Here, Z is an element placeholder, not a supplied atomic number. The choices are:

How do you identify the first two emissions?

The first emission is alpha, and the second is beta-minus, establishing D and C respectively. Follow the official solution: check both mass number and atomic number at each arrow. Mass number counts protons and neutrons together; atomic number counts protons alone.

Use these daughter-minus-parent changes, with mass number listed first and atomic number second:

Alpha:

(Δ A, Δ Z_{atomic}) = (- 4, - 2)

Beta-minus:

(Δ A, Δ Z_{atomic}) = (0, + 1)

Gamma:

(Δ A, Δ Z_{atomic}) = (0, 0)

An alpha particle is a helium nucleus with positive charge. Its nuclear symbol and charge are:

α =_{2}^{4} H e, q_{α} = + 2 e

For uranium-238 becoming thorium-234, calculate both changes:

234 - 238 = - 4, 90 - 92 = - 2

These match alpha emission. Its positive charge means it deflects towards the negatively charged plate, so D is true. $x_{1} = α$

For thorium-234 becoming protactinium-234, the changes are:

234 - 234 = 0, 91 - 90 = + 1

This is beta-minus decay: a neutron converts into a proton, with an electron and an antineutrino emitted. The proton count rises by one, but the total number of protons and neutrons stays unchanged. $x_{2} = β^{-}$

Thus C is true. The unchanged mass number alone does not establish this answer; the increased atomic number does.

How does the last arrow identify the missing uranium nucleus?

The known final daughter is thorium-230, with atomic number 90, while the preceding intermediate has mass number 234. Working backwards from this daughter identifies the intermediate as uranium-234, without guessing from a remembered decay chain.

Start with the final mass-number change: $230 - 234 = - 4$

Among the decay rules being used, this identifies the fourth emission as alpha. Reverse the alpha changes to recover the parent: $A_{parent} = 230 + 4 = 234$

Z_{atomic,parent} = 90 + 2 = 92

Atomic number 92 identifies uranium. Therefore, the intermediate is:

^{234} Z =_{92}^{234} U

Uranium-234 and the starting uranium-238 both have atomic number 92 but different mass numbers. That makes them isotopes, so B is true.

Now return to the third arrow with the missing atomic number filled in:

_{91}^{234} P a ⟶_{92}^{234} U

234 - 234 = 0, 92 - 91 = + 1 \Rightarrow x_{3} = β^{-}

Gamma emission cannot raise atomic number, so A is false. Finally, verify both numbers for the last arrow rather than stopping at the mass check:

_{92}^{234} U ⟶_{90}^{230} T h

230 - 234 = - 4, 90 - 92 = - 2 \Rightarrow x_{4} = α

Why are B, C and D the correct options?

Select B, C and D. The complete emission sequence is:

x_{1} = α, x_{2} = β^{-}, x_{3} = β^{-}, x_{4} = α

Audit each claim separately before submitting a multiple-correct answer:

A, false: The third transition raises atomic number, ruling out gamma.
B, true: The intermediate is uranium-234.
C, true: The second transition is beta-minus.
D, true: Alpha particles carry positive charge and deflect towards a negative plate.

As a brief consistency check, two alpha emissions and two beta-minus emissions give: $A_{final} = 238 - 8 = 230$

Z_{atomic,final} = 92 - 4 + 2 = 90

Why is choosing gamma for the third emission a method error?

Choosing A can result from checking only mass number. The faulty inference is that the third transition keeps mass number at 234, so it must emit gamma radiation. This fails because both beta-minus and gamma emission leave mass number unchanged; atomic number must also be checked.

The actual atomic number rises: $91 ⟶ 92$

This requires beta-minus emission and rules out gamma. If the third emission were gamma, the intermediate would retain atomic number 91, and the final alpha emission would then give:

Z_{atomic,final} = 91 - 2 = 89 \neq 90

That cannot produce the stated thorium daughter. Record two entries at every arrow: mass-number change and atomic-number change, even when the first entry appears to suggest an answer.

How can you practise this method on three related questions?

Practise emission counting, gamma decay and backward identification separately. These are original related practice questions, not additional verified PYQs; each solution follows its question.

Question 1: An unspecified sequence containing only alpha and beta-minus emissions converts uranium-238 into uranium-234, both with atomic number 92. How many of each emission occur?

Define the counts, then use mass number first and atomic number second:

a = alpha count, b = beta-minus count

238 - 4 a = 234 \Rightarrow a = 1

92 - 2 a + b = 92 \Rightarrow b = 2

Answer: one alpha and two beta-minus emissions. This determines their counts, not their order.

Question 2: An excited nucleus with mass number 234 and atomic number 92 emits one gamma photon. What are the daughter’s two numbers?

Answer: 234 and 92. Gamma emission reduces nuclear energy without changing either number.

A_{daughter} = 234, Z_{atomic,daughter} = 92

Question 3: A nucleus emits one alpha particle to form thorium-234, whose atomic number is 90. Identify the parent and the emitted particle’s deflection direction near charged plates.

Reverse both alpha changes:

A_{parent} = 234 + 4 = 238, Z_{atomic,parent} = 90 + 2 = 92

Parent =_{92}^{238} U

The positively charged alpha particle deflects towards the negative plate. Cover the solutions and write both number checks for each drill before checking your answers.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What are the correct options for the radioactive decay series JEE 2019 question?

B, C and D are correct in JEE Advanced 2019, Paper 1, Chemistry, Q.8. The four emissions are alpha, beta-minus, beta-minus and alpha, in that order. The missing intermediate is uranium-234.

How do you identify the missing nucleus in the decay series?

Work backwards from thorium-230, whose atomic number is 90. The final alpha emission reduces mass number by 4 and atomic number by 2, so its parent has mass number 234 and atomic number 92. The missing nucleus is therefore uranium-234.

Why is the third emission beta-minus and not gamma?

The third transition converts protactinium-234 into uranium-234, increasing atomic number from 91 to 92 while keeping mass number unchanged. This matches beta-minus decay. Gamma emission changes neither number, so unchanged mass number alone cannot identify it.

Why does the first emitted particle move towards the negative plate?

The first transition reduces mass number from 238 to 234 and atomic number from 92 to 90, identifying alpha emission. An alpha particle is a helium nucleus with charge +2e, so it deflects towards the negatively charged plate.