Q-value of Nuclear Reaction NEET 2026: 3726 keV Solved

What is the answer to the uranium alpha-decay Q-value question?

The correct answer to the Q-value of nuclear reaction NEET 2026 question is A, 3726 keV. This Physics question belongs to Atoms and Nuclei and is tagged medium by the question bank.

A uranium-238 nucleus decays into thorium-234 and helium-4. Using the supplied masses, calculate the reaction’s Q-value, the total energy released, in keV.

^{238} U \to^{234} T h +^{4} H e

Use the following masses for uranium, thorium and helium, respectively. Keep their decimal values intact throughout the subtraction.

m_{U} = 238.050 u, m_{T h} = 234.043 u, m_{H e} = 4.003 u

The isotope mass numbers are not substitutes for these masses. The options are close enough that rounding before the unit conversion can change which answer you select.

Which masses should you subtract to find the mass difference?

$Q = (m_{i} - m_{f}) c^{2}$ Subtract the combined mass of every product from the initial uranium mass. The final mass includes both thorium and helium, so add their masses before subtracting. This is the mass bookkeeping used in the official solution. $m_{i} = 238.050 u$

m_{f} = 234.043 u + 4.003 u = 238.046 u

Δ m = m_{i} - m_{f} = 238.050 u - 238.046 u = 0.004 u

Subtracting only the thorium mass would count the helium product as though it had disappeared into energy. Helium remains a product with its own rest mass, so omitting it overstates the mass converted into energy.

Keep all the supplied decimal values intact. The integers in the isotope names count nucleons; they are not the given isotope masses in atomic mass units. Using those integers would erase the small mass difference that this question asks you to convert into energy.

How do you convert the mass difference into 3726 keV?

The mass difference gives 3.726 MeV, which converts to 3726 keV. Retain every computed digit until you have expressed the answer in the requested unit.

The official solution uses this mass–energy conversion. It gives the energy equivalent of one atomic mass unit:

1 u \times c^{2} = 931.5 M e V

Q = 0.004 u \times 931.5 \frac{M e V}{u}

$Q = 3.726 M e V$ The mass units cancel, leaving energy in MeV. Convert to keV without rounding:

1 M e V = 1000 k e V

Q = 3.726 \times 1000 k e V = 3726 k e V

Correct answer: A, 3726 keV. This matches the unrounded result in the unit the question requests.

Do not multiply by the square of the speed of light again. That factor is already included in the conversion from atomic mass units to MeV. Applying it a second time would give incorrect units as well as an incorrect value.

What does the positive Q-value tell you?

The positive result means that this reaction releases energy, rather than requiring a net input of reaction energy. The initial rest mass exceeds the combined final rest mass:

238.050 u > 238.046 u, Q > 0

The 3726 keV is the total reaction energy release, not automatically the helium nucleus’s kinetic energy alone. The thorium daughter recoils and carries kinetic energy too.

The intermediate answer, 3.726 MeV, checks the conversion: expressing the same energy in keV increases its numerical value by a factor of 1000. Repeat the sequence: add product masses, subtract from reactant mass, multiply by 931.5, then convert units.

How can premature rounding produce option B, 3730 keV?

Rounding 3.726 MeV to 3.73 MeV before converting produces 3730 keV, option B. The multiplication is arithmetically correct, but it uses a value whose precision has already been reduced.

Keep the computed digits: $3.726 \times 1000 = 3726$
Round to two decimal places first:

3.726 M e V \overset{rounding}{\to} 3.73 M e V

$3.73 \times 1000 = 3730$ Two decimal places in MeV discard information needed to distinguish the supplied keV options. The rounding shifts the answer by 4 keV, exactly the separation between options A and B. Converting units after rounding cannot recover those discarded digits.

This is a reproducible route to a wrong answer, not evidence of what the examiner intended or which option students usually choose. Carry the computed digits through the conversion and match the unrounded result to the options.

Which related practice questions check the sign and meaning of Q-value?

Reverse the reaction to check the sign; consider recoil to check how the released energy is shared. The following are original related practice questions based on the supplied reaction, not additional verified PYQs.

What is the Q-value if the reaction runs in reverse?

Related practice question 1: Using the same masses, find the Q-value for the reverse reaction shown below. Worked answer: −3726 keV, because the initial and final masses exchange places.

^{234} T h +^{4} H e \to^{238} U

m_{i} = (234.043 + 4.003) u = 238.046 u

m_{f} = 238.050 u

Δ m = 238.046 u - 238.050 u = - 0.004 u

Q = - 0.004 u \times 931.5 \frac{M e V}{u} = - 3.726 M e V = - 3726 k e V

Reversing the reaction reverses the sign of its Q-value. Do not treat its magnitude as a general laboratory threshold-energy formula; laboratory thresholds also depend on momentum conservation and the reaction setup.

Must helium receive all 3726 keV if uranium starts at rest?

Related practice question 2: If the uranium parent is initially at rest, must helium alone receive all the released energy as kinetic energy? Worked answer: No. The initial total momentum is zero, so the thorium daughter must recoil with momentum opposite to helium’s.

The released energy is shared between the two products. Do not assign all 3726 keV to helium.

Before checking your next answer, identify the quantity requested: total reaction Q-value or one product’s kinetic energy. That distinction decides whether the mass-difference calculation alone finishes the problem.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

Keep going with How to Study Laws of Motion NEET: FBDs and Friction.

Frequently asked questions

How do you calculate the Q-value of uranium-238 alpha decay?

Use Q = (initial mass − total final mass)c², including both thorium and helium as products. With the supplied masses, the mass difference is 238.050 − (234.043 + 4.003) = 0.004 u. Multiplying by 931.5 MeV/u gives 3.726 MeV, or 3726 keV: option A.

Why do we multiply the mass difference by 931.5?

One atomic mass unit has an energy equivalent of 931.5 MeV, so multiplying the mass difference in u by 931.5 MeV/u gives Q in MeV. The factor c² is already included in this conversion and must not be applied again. Multiply the MeV result by 1000 to express it in keV.

Why is 3730 keV wrong in this Q-value question?

Using the supplied masses and conversion factor gives 3.726 MeV, which equals 3726 keV. Rounding first to 3.73 MeV produces 3730 keV, option B. Keep the computed digits until after converting to the requested unit.

Does the alpha particle receive all 3726 keV?

No: 3726 keV is the total reaction energy release, not the alpha particle's kinetic energy alone. If uranium starts at rest, the thorium daughter recoils with momentum opposite to the alpha particle's. Both products therefore carry kinetic energy.

Q-value of Nuclear Reaction NEET 2026: Uranium Alpha Decay