In the decay `Cu^64 rarr Ni^64 + e^+ + v`,
the maximum kinetic energy carried by the netrino which was emitted together with a positron of kinetic energy 0.150 meV? (a) what is the energy of the neutrino which was emitted togather with a positron of kinetic energy 0.150 MeV ? .(b)What is the momentum of this neutrino in kg m `s^(-1)`?Use the formula applicable to photon.

Explanation

To solve the problem, we will go through the steps systematically.

### Given:
- The decay process is: \( \text{Cu}^{64} \rightarrow \text{Ni}^{64} + e^+ + \nu \)
- The kinetic energy of the emitted positron \( K = 0.150 \, \text{MeV} \)

### (a) Finding the energy of the neutrino

1. **Understanding Energy Conservation**:
In the decay process, the total energy before and after the decay must be conserved. The energy of the neutrino (\( E_{\nu} \)) can be calculated using the kinetic energy of the positron.

2. **Using the Energy Conservation Equation**:
The total energy released in the decay is shared between the positron and the neutrino. Since the neutrino is very light, we can consider its kinetic energy to be equal to its total energy.
\[ E_{\nu} = K - 0 \]
Since the rest mass energy of the neutrino is negligible, we can assume that all the energy goes to the kinetic energy of the neutrino.

3. **Calculating the Energy of the Neutrino**:
\[ E_{\nu} = 0.150 \, \text{MeV} \]

### (b) Finding the momentum of the neutrino

1. **Using the formula for momentum of a photon**:
The momentum (\( p \)) of a particle can be calculated using the formula:
\[ p = \frac{E}{c} \]
where \( E \) is the energy of the particle and \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

2. **Substituting the values**:
\[ p = \frac{0.150 \, \text{MeV}}{c} \]
To convert MeV to Joules, we use the conversion \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \):
\[ p = \frac{0.150 \times 1.6 \times 10^{-13} \, \text{J}}{3 \times 10^8 \, \text{m/s}} \]

3. **Calculating the momentum**:
\[ p = \frac{0.240 \times 10^{-13} \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 0.8 \times 10^{-22} \, \text{kg m/s} \]
\[ p = 8.0 \times 10^{-23} \, \text{kg m/s} \]

### Final Answers:
(a) The energy of the neutrino is \( 0.150 \, \text{MeV} \).

(b) The momentum of the neutrino is \( 8.0 \times 10^{-23} \, \text{kg m/s} \).