The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 α particles is \_\_\_\_ kHz . [mass of He nucleus = 4.002amu, 1amu = 1.66 × 10-27 kg, h=6.6 × 10-34 J. s and c=3 × 10^{8}\,{m/s} ] [22 Jan 2026 Shift 1]

Correct Answer is : 14.94×101914.94 × 10¹9}14.94×1019

Correct Answer is : 14.94×101914.94 × 10¹9}14.94×1019

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Atoms and Nuclei Part 3

Section: Physics

Question : 22 of 98

Marks: +1, -0

4 \alpha

\_\_\_\_

\times 10^{8}\,\text{m/s}

Solution:

Concept:

Mass defect is converted to energy using Einstein's equation

E = \Delta m c^2

. This energy is supplied by a photon of frequency

v = E/h

Explanation:

Compute the mass defect

\Delta m

Initial mass =

15.348

amu.

Final mass (4 He nuclei) =

4 \times 4.002

amu =

16.008

amu.

\Delta m = 16.008 - 15.348 = 0.66

amu.

Convert to kg:

\Delta m = 0.66 \times (1.66 \times 10^{-27}) = 1.0956 \times 10^{-27}

kg.

Energy released:

E = \Delta m c^2 = (1.0956 \times 10^{-27}) \times (3 \times 10^8)^2 = 1.0956 \times 10^{-27} \times 9 \times 10^{16} = 9.8604 \times 10^{-11}

Photon frequency:

v = E/h = (9.8604 \times 10^{-11}) / (6.6 \times 10^{-34}) = 1.494 \times 10^{23}

Hz.

Convert to kHz:

v = 1.494 \times 10^{23} \times 10^{-3} = 1.494 \times 10^{20}

kHz =

14.94 \times 10^{19}

kHz.

Answer:

14.94 \times 10^{19}

kHz, which corresponds to option D.

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