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Calculate the energy retained by an beta particle, if its incoming energy is 3.9 keV and it ionizes an L-shell electron of an oxygen atom (binding energy 42 eV) and the ionized electron has 200 eV of kinetic energy.

Calculate the energy retained by an beta particle, if its incoming energy is 3.9 keV and it ionizes an L-shell electron of an oxygen atom (binding energy 42 eV) and the ionized electron has 200 eV of kinetic energy.

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**Problem: Energy Retention by a Beta Particle** Calculate the energy retained by a beta particle if its incoming energy is 3.9 keV and it ionizes an L-shell electron of an oxygen atom (binding energy 42 eV). The ionized electron has 200 eV of kinetic energy. **Solution:** To find the energy retained by the beta particle, subtract the energy used in the ionization process (sum of the binding energy and the kinetic energy of the ionized electron) from the incoming energy: \[ \text{Energy retained} = \text{Incoming energy} - (\text{Binding energy} + \text{Kinetic energy}) \] \[ \text{Energy retained} = 3.9 \text{ keV} - (42 \text{ eV} + 200 \text{ eV}) \] Convert keV to eV for consistency: \[ 3.9 \text{ keV} = 3900 \text{ eV} \] Now calculate: \[ \text{Energy retained} = 3900 \text{ eV} - 242 \text{ eV} = 3658 \text{ eV} \] Thus, the energy retained by the beta particle is 3658 eV.
Transcribed Image Text:**Problem: Energy Retention by a Beta Particle** Calculate the energy retained by a beta particle if its incoming energy is 3.9 keV and it ionizes an L-shell electron of an oxygen atom (binding energy 42 eV). The ionized electron has 200 eV of kinetic energy. **Solution:** To find the energy retained by the beta particle, subtract the energy used in the ionization process (sum of the binding energy and the kinetic energy of the ionized electron) from the incoming energy: \[ \text{Energy retained} = \text{Incoming energy} - (\text{Binding energy} + \text{Kinetic energy}) \] \[ \text{Energy retained} = 3.9 \text{ keV} - (42 \text{ eV} + 200 \text{ eV}) \] Convert keV to eV for consistency: \[ 3.9 \text{ keV} = 3900 \text{ eV} \] Now calculate: \[ \text{Energy retained} = 3900 \text{ eV} - 242 \text{ eV} = 3658 \text{ eV} \] Thus, the energy retained by the beta particle is 3658 eV.
Expert Solution
Step 1

Given:

  • Incoming Energy(Eincoming/Initial)=3.9KeV=3900eV
  • Binding energy=42eV
  • k.E of Ionized electron=200eV

To find:

  • Energy Retained by an beta particles.
steps

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