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MCQEasyJEE 2026Atomic Mass & Binding Energy

JEE Physics 2026 Question with Solution

The minimum frequency of photon required to break a particle of mass 15.34815.348 amu into 44 particles is _____ kHz.

[Mass of He nucleus =4.002=4.002 amu, 11 amu =1.66×10−27=1.66\times10^{-27} kg, h=6.6×10−34h=6.6\times10^{-34} J⋅\cdots and c=3×108c=3\times10^8 m/s]

  • A

    9×10199\times10^{19}

  • B

    9×10209\times10^{20}

  • C

    14.94×102014.94\times10^{20}

  • D

    14.94×101914.94\times10^{19}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Initial particle mass is 15.34815.348 amu. It breaks into 44 He nuclei of mass 4.0024.002 amu each. Also, 11 amu =1.66×10−27=1.66\times10^{-27} kg, h=6.6×10−34h=6.6\times10^{-34} J⋅\cdots, and c=3×108c=3\times10^8 m/s.

Find: The minimum frequency of the photon required.

Using mass defect, the required energy is

E=Δmc2E=\Delta mc^2

Initial mass:

mi=15.348 amum_i = 15.348\text{ amu}

Final mass:

mf=4×4.002=16.008 amum_f = 4\times4.002 = 16.008\text{ amu}

Therefore, mass defect:

Δm=mf−mi=0.660 amu\Delta m = m_f - m_i = 0.660\text{ amu}

Convert this mass defect into kilogram:

Δm=0.660×1.66×10−27=1.0956×10−27 kg\Delta m = 0.660 \times 1.66\times10^{-27} =1.0956\times10^{-27}\text{ kg}

Energy required:

E=Δmc2=1.0956×10−27(3×108)2=9.86×10−11 JE=\Delta mc^2 =1.0956\times10^{-27}(3\times10^8)^2 =9.86\times10^{-11}\text{ J}

Now use the photon relation:

E=hνE = h\nu

So,

ν=Eh\nu = \frac{E}{h}

Substituting values,

ν=9.86×10−116.6×10−34=1.494×1020 Hz\nu=\frac{9.86\times10^{-11}}{6.6\times10^{-34}} =1.494\times10^{20}\text{ Hz} =14.94×1019 Hz=14.94\times10^{19}\text{ Hz}

Therefore, the minimum photon frequency is 14.94×1019 Hz14.94\times10^{19}\text{ Hz} and the correct option is D.

Mass Defect to Photon Frequency

Given: The nucleus has mass 15.34815.348 amu and the final products are 44 helium nuclei each of mass 4.0024.002 amu.

Find: The threshold photon frequency.

The final total mass is greater than the initial mass, so energy must be supplied to make the reaction possible. That supplied energy comes from the photon.

First calculate the total final mass:

4×4.002=16.008 amu4\times 4.002 = 16.008\text{ amu}

Then compare with the initial mass:

16.008−15.348=0.660 amu16.008 - 15.348 = 0.660\text{ amu}

This excess mass corresponds to required energy:

E=Δmc2E = \Delta mc^2

With

Δm=0.660×1.66×10−27 kg\Delta m = 0.660 \times 1.66\times10^{-27}\text{ kg}

we get

E=9.86×10−11 JE = 9.86\times10^{-11}\text{ J}

For a photon,

E=hνE=h\nu

Hence,

ν=9.86×10−116.6×10−34=1.494×1020 Hz\nu = \frac{9.86\times10^{-11}}{6.6\times10^{-34}} = 1.494\times10^{20}\text{ Hz}

Therefore, the required frequency is 1.494×1020 Hz1.494\times10^{20}\text{ Hz}, that is 14.94×1019 Hz14.94\times10^{19}\text{ Hz}. So the correct option is D.

Common mistakes

  • Using Δm=mi−mf\Delta m = m_i - m_f here gives a negative value and leads to the wrong physical interpretation. Since the final mass is larger, the photon must supply energy, so use the magnitude of the mass increase to compute the required energy.

  • Forgetting to multiply the helium nucleus mass by 44 is incorrect because the particle breaks into 44 particles. First find the total final mass 4×4.0024\times4.002 amu, then compare it with the initial mass.

  • Confusing photon relations such as E=hc/λE=hc/\lambda and E=hνE=h\nu can cause unnecessary errors. Since the question asks for frequency directly, use E=hνE=h\nu after finding the energy from mass defect.

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