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:
For nuclear reactions induced by photons, the minimum photon energy required equals the mass defect energy: \[ E = \Delta m\,c^2 \] The photon energy is also given by: \[ E = h\nu \] Hence, \[ h\nu = \Delta m\,c^2 \]
Step 1: Calculate the mass defect. Initial mass: \[ m_i = 15.356\,\text{amu} \] Final mass (four \(\alpha\)-particles): \[ m_f = 4 \times 4.004 = 16.016\,\text{amu} \] Mass defect: \[ \Delta m = m_f - m_i = 16.016 - 15.356 = 0.660\,\text{amu} \]
Step 2: Convert mass defect into energy. Using: \[ 1\,\text{amu} = 1.66\times10^{-27}\,\text{kg}, \quad c = 3\times10^8\,\text{m s}^{-1} \] \[ \Delta m = 0.660 \times 1.66\times10^{-27} = 1.096\times10^{-27}\,\text{kg} \] \[ E = \Delta m\,c^2 = 1.096\times10^{-27}\times (3\times10^8)^2 \] \[ E = 9.864\times10^{-11}\,\text{J} \]
Step 3: Find the frequency of the photon. \[ \nu = \frac{E}{h} = \frac{9.864\times10^{-11}}{6.6\times10^{-34}} \] \[ \nu = 1.49\times10^{23}\,\text{Hz} \]
Step 4: Convert Hz to kHz. \[ \nu = 1.49\times10^{20}\,\text{kHz} \approx 14.9\times10^{19}\,\text{kHz} \] \[ \boxed{\nu = 14.9\times10^{19}\,\text{kHz}} \]