Explanation

This problem is based on mass–energy equivalence and is a standard application of nuclear physics for JEE Main, JEE Advanced and IIT JEE.

Initial mass of the particle:

$$ m_i = 15.348 \ \text{amu} $$

Final mass after breaking into 4 alpha particles:

$$ m_f = 4 \times 4.002 = 16.008 \ \text{amu} $$

Mass defect required:

$$ \Delta m = m_f - m_i = 16.008 - 15.348 = 0.660 \ \text{amu} $$

Convert mass defect into kg:

$$ \Delta m = 0.660 \times 1.66 \times 10^{-27} = 1.0956 \times 10^{-27} \ \text{kg} $$

Energy required:

$$ E = \Delta m c^2 $$

$$ E = 1.0956 \times 10^{-27} \times (3 \times 10^8)^2 $$

$$ E = 9.86 \times 10^{-11} \ \text{J} $$

Photon energy is also given by:

$$ E = h \nu $$

$$ \nu = \frac{E}{h} $$

$$ \nu = \frac{9.86 \times 10^{-11}}{6.63 \times 10^{-34}} $$

$$ \nu = 1.494 \times 10^{23} \ \text{Hz} $$

Converting into kHz:

$$ \nu = 1.494 \times 10^{20} \ \text{kHz} $$

Therefore, the minimum required frequency is:

$14.94 \times 10^{19}$ kHz

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