Question

The minimum frequency of photon required to break a particle of mass $15.348$ amu into $4$ particles is __________ kHz.
[Mass of He nucleus $=4.002$ amu, $1$ amu $=1.66\times10^{-27}$ kg, $h=6.6\times10^{-34}$ J$\cdot$s and $c=3\times10^8$ m/s]

• A. $9\times10^{19}$
• B. $9\times10^{20}$
• C. $14.94\times10^{20}$
• D. $14.94\times10^{19}$

Hint:

For nuclear reactions, required photon energy is calculated using mass defect: $E=\Delta mc^2$.

Answer 1

The Correct Option is D

Step 1: Find mass defect.
Initial mass: \[ m_i = 15.348\text{ amu} \] Final mass: \[ m_f = 4\times4.002 = 16.008\text{ amu} \] Mass defect: \[ \Delta m = m_f - m_i = 0.660\text{ amu} \] Step 2: Convert mass defect into energy.
\[ \Delta m = 0.660 \times 1.66\times10^{-27} =1.0956\times10^{-27}\text{ kg} \] Energy required: \[ E=\Delta mc^2 =1.0956\times10^{-27}(3\times10^8)^2 =9.86\times10^{-11}\text{ J} \] Step 3: Use photon energy relation.
\[ E = h\nu \Rightarrow \nu = \frac{E}{h} \] \[ \nu=\frac{9.86\times10^{-11}}{6.6\times10^{-34}} =1.494\times10^{20}\text{ Hz} \] \[ =14.94\times10^{19}\text{ Hz} \] Final Answer: $\boxed{14.94\times10^{19}\text{ Hz}}$