Question

A photon is incident on a particle having mass \(m = 15.356\,\text{amu}\). What should be the frequency of the photon so that the particle of mass \(m\) breaks into four \(\alpha\)-particles? (Given: \(m_\alpha = 4.004\,\text{amu}\); \(h = 6.6\times10^{-34}\,\text{J s}\))

• A. \(14.9\times10^{19}\,\text{kHz}\)
• B. \(12.9\times10^{19}\,\text{kHz}\)
• C. \(9.9\times10^{19}\,\text{kHz}\)
• D. \(10.9\times10^{19}\,\text{kHz}\)

Hint:

For photon-induced nuclear reactions:
Required photon energy equals mass defect energy
Always check whether final mass is greater or smaller than initial mass
Convert amu carefully into SI units before calculation

Answer 1

The Correct Option is A

Concept:  
The energy required to break the particle into four α-particles comes from the energy of the photon. The total energy provided by the photon is equal to the difference in the mass-energy before and after the particle breaks apart. 
Step 1: Use the mass-energy equivalence formula 
According to Einstein’s equation, energy and mass are related by: \[ E = \Delta m \cdot c^2 \] where \( \Delta m \) is the mass defect, and \( c \) is the speed of light. 
Step 2: Calculate the mass defect 
The mass defect \( \Delta m \) is the difference between the mass of the original particle and the total mass of the four α-particles: \[ \Delta m = m - 4 \cdot m_{\alpha} \] Substituting the given values: \[ \Delta m = 15.356 \, \text{amu} - 4 \cdot 4.004 \, \text{amu} \] \[ \Delta m = 15.356 \, \text{amu} - 16.016 \, \text{amu} \] \[ \Delta m = -0.66 \, \text{amu} \] 
Step 3: Convert the mass defect into kilograms 
1 amu = \(1.660539 \times 10^{-27} \, \text{kg}\), so: \[ \Delta m = -0.66 \, \text{amu} \times 1.660539 \times 10^{-27} \, \text{kg/amu} \] \[ \Delta m = -1.095 \times 10^{-27} \, \text{kg} \] 
Step 4: Calculate the energy 
The energy \( E \) corresponding to the mass defect is: \[ E = \Delta m \cdot c^2 \] where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light. Substituting the values: \[ E = (-1.095 \times 10^{-27} \, \text{kg}) \cdot (3 \times 10^8 \, \text{m/s})^2 \] \[ E = -1.095 \times 10^{-27} \, \text{kg} \cdot 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \] \[ E = -9.855 \times 10^{-11} \, \text{J} \] 
Step 5: Relating energy to the frequency of the photon 
The energy of the photon \( E_{\text{photon}} \) is related to its frequency \( f \) by Planck's equation: \[ E_{\text{photon}} = h \cdot f \] where \( h = 6.626 \times 10^{-34} \, \text{J·s} \) is Planck's constant. Solving for \( f \): \[ f = \frac{E_{\text{photon}}}{h} \] Substituting the energy required for the particle to break: \[ f = \frac{9.855 \times 10^{-11}}{6.626 \times 10^{-34}} \, \text{Hz} \] \[ f = 1.49 \times 10^{19} \, \text{Hz} \] 
Step 6: Convert to kHz \[ f = 1.49 \times 10^{19} \, \text{Hz} = 14.9 \times 10^{18} \, \text{Hz} = 14.9 \times 10^{19} \, \text{kHz} \] 
Final Answer: \[ \boxed{14.9 \times 10^{19} \, \text{kHz}} \]