Atoms and Nuclei NEET 2026: Uranium Decay Q-Value Explained

What is the answer to the uranium-decay question in Atoms and Nuclei NEET 2026?

Option A, 3726 keV, is correct. This Atoms and Nuclei NEET 2026 question comes from NEET UG 2026, re-held on 2026-06-21, Code-50. “Medium” is the question bank’s difficulty classification, not an official NTA rating.

Find the energy released per reaction, in keV, when uranium-238 transforms into thorium-234 and helium-4. Use the supplied masses and conversion factor:

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

m (U) = 238.050 u, m (T h) = 234.043 u, m (H e) = 4.003 u

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

The gap between options A and B is only 4 keV. Preserve the given conversion factor rather than replacing it with a remembered approximation.

Which masses should you subtract in step 1?

Q = [mass of reactants - mass of products] c^{2}

Subtract the combined mass of both products from the uranium mass. This follows the official solution’s mass-defect method. Put thorium and helium together inside brackets before subtracting:

Δ m = m (U) - [m (T h) + m (H e)]

First, add the product masses:

m (products) = (234.043 + 4.003) u = 238.046 u

Now subtract, keeping the decimal places aligned:

\begin{matrix} 238.050 \\ - 238.046 \\ 0.004 \end{matrix} Δ m = (238.050 - 238.046) u = 0.004 u

Mass number is not the supplied mass in atomic mass units. The following equality checks conservation of nucleon number: $238 = 234 + 4$

It does not show that the supplied masses add exactly, or that the released energy is zero. Nucleon count is conserved while rest mass can decrease.

Use the stated masses directly. Bringing in external mass values would change the numerical problem instead of solving the one asked.

How do you convert the mass difference into keV in step 2?

The mass difference gives 3.726 MeV, which converts to 3726 keV. Multiply by the supplied mass-to-energy factor first, then convert MeV to keV. There is no need to convert through kilograms or joules.

Q = (0.004 u) (931.5 \frac{M e V c^{- 2}}{u}) c^{2} = 0.004 \times 931.5 M e V

The atomic mass units cancel, as do the powers of the speed of light:

\frac{u}{u} = 1, c^{- 2} c^{2} = 1

Check the multiplication before placing the decimal: $931.5 \times 4 = 3726$

Q = \frac{3726}{1000} M e V = 3.726 M e V

Then convert the energy unit:

1 M e V = 1000 k e V

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

Final answer: option A, 3726 keV. The division by 1000 belongs to the decimal multiplication; the later multiplication by 1000 changes the energy unit. Neither step changes the supplied conversion factor.

How do you check the sign and meaning in step 3?

The initial mass is larger than the total final mass, so the reaction’s Q-value is positive. Positive Q means energy is released.

238.050 u > 238.046 u \Rightarrow Q > 0

The 3726 keV is the total reaction Q-value, not automatically the kinetic energy of the helium nucleus alone. The thorium nucleus can recoil and carry kinetic energy too.

Use an approximate conversion factor only for a rough scale check:

Q_{estimate} \approx (0.004 u) (1000 \frac{M e V}{u}) = 4 M e V

This estimate checks the scale, not the option. Use the exact supplied factor to distinguish nearby answers.

The reusable sequence is: add product masses, subtract from reactant mass, convert to energy, check the requested unit.

Why does changing the conversion factor give option B?

Option B can be reproduced by keeping the correct mass difference but replacing the supplied 931.5 with 932.5. This is an unjustified alteration, not legitimate rounding. It reproduces the wrong answer without making the calculation valid.

The incorrect substitution gives:

Q_{wrong} = 0.004 \times 932.5 M e V = 3.730 M e V = 3730 k e V

The change in the energy-conversion factor is:

(932.5 - 931.5) M e V / u = 1 M e V / u

Its effect on the answer is:

δ Q = (0.004 u) (1 M e V / u) = 0.004 M e V = 4 k e V

That accounts exactly for the gap between options A and B. Rounding reduces numerical precision; it does not justify changing 931.5 to 932.5.

Use the supplied conversion factor unchanged and postpone rounding until the calculation is complete. This arithmetic explains B, but it does not establish how options C and D were constructed.

What three related questions can you solve from the same reaction?

Use the same masses to check the reverse reaction, recover the mass difference from energy, and compare combined binding energies. These are original practice extensions, not additional verified NEET PYQs.

What is the Q-value of the formal reverse reaction in Question 1?

The Q-value is negative 3726 keV because reversing the reaction reverses the mass subtraction. For the formal reverse reaction:

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

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

Q = - 0.004 \times 931.5 M e V = - 3726 k e V

The formal reverse process is endothermic: it requires energy. Do not equate the magnitude of this Q-value with a laboratory threshold energy, which also depends on reaction kinematics.

What mass difference corresponds to 3.726 MeV in Question 2?

The corresponding mass difference is 0.004 u. Divide the energy by the supplied mass-to-energy conversion factor, rather than multiplying again:

Δ m = \frac{Q}{c^{2}} = \frac{3.726}{931.5} u = 0.004 u

How does the products’ combined binding energy compare with uranium’s in Question 3?

The products’ combined binding energy is greater by 3.726 MeV. Total proton and neutron numbers are conserved, so the separate-nucleon reference masses cancel when comparing the initial and final binding energies:

Q = B E (T h) + B E (H e) - B E (U)

B E (T h) + B E (H e) = B E (U) + 3.726 M e V

This compares the sum of the products’ binding energies with uranium’s binding energy. It does not mean that each product individually has greater total binding energy than uranium.

Cover the calculations and redo the three extensions. For each, write the subtraction order or conversion direction before touching the numbers.

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).

For a worked example of the same idea, see How to Study Sexual Reproduction in Flowering Plants for NEET.

Frequently asked questions

How do you calculate the energy released in the uranium-238 decay question?

Subtract both product masses from the uranium mass: 238.050 − (234.043 + 4.003) = 0.004 u. Multiply by the supplied factor to get 0.004 × 931.5 = 3.726 MeV, then convert to 3726 keV. The correct answer is option A.

Why is energy released if 238 equals 234 plus 4?

The equality 238 = 234 + 4 shows conservation of nucleon number, not equality of the supplied masses. The products have a combined mass of 238.046 u, which is 0.004 u less than the uranium mass. This decrease in rest mass gives a positive Q-value, meaning energy is released.

Why is 3730 keV the wrong answer?

Using 932.5 instead of the supplied 931.5 gives 0.004 × 932.5 = 3.730 MeV, or 3730 keV. That substitution is an unjustified change, not legitimate rounding. Keep the supplied conversion factor unchanged to obtain 3726 keV.

Is 3726 keV the kinetic energy of the alpha particle?

No, 3726 keV is the total reaction Q-value calculated from the supplied masses. The thorium nucleus can recoil and carry kinetic energy too, so the total released energy cannot automatically be assigned to the alpha particle alone.

Atoms and Nuclei NEET 2026: Uranium Decay Q-Value