An electron and a photon each have a wavelength of 2.00nm. Find (i) their momenta (ii) the energy of the photon and (iii) the kinetic energy of electron. `h=6.63xx10^(-34)Js`

Explanation

To solve the problem step by step, we will find the momentum of both the electron and the photon, the energy of the photon, and the kinetic energy of the electron.

### Step 1: Calculate the momentum of the electron

The momentum \( P_E \) of an electron can be calculated using the de Broglie wavelength formula:

\[ P_E = \frac{h}{\lambda_E} \]

Where:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( \lambda_E = 2.00 \, \text{nm} = 2.00 \times 10^{-9} \, \text{m} \) (wavelength of the electron)

Substituting the values:

\[ P_E = \frac{6.63 \times 10^{-34}}{2.00 \times 10^{-9}} = 3.315 \times 10^{-25} \, \text{kg m/s} \]

### Step 2: Calculate the momentum of the photon

The momentum \( P_P \) of a photon can also be calculated using the same formula since the wavelength is the same for both:

\[ P_P = \frac{h}{\lambda_P} \]

Where \( \lambda_P = 2.00 \, \text{nm} = 2.00 \times 10^{-9} \, \text{m} \).

Since \( \lambda_E = \lambda_P \):

\[ P_P = P_E = 3.315 \times 10^{-25} \, \text{kg m/s} \]

### Step 3: Calculate the energy of the photon

The energy \( E \) of the photon can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( c = 3.00 \times 10^{8} \, \text{m/s} \) (speed of light)
- \( \lambda = 2.00 \, \text{nm} = 2.00 \times 10^{-9} \, \text{m} \)

Substituting the values:

\[ E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{2.00 \times 10^{-9}} = 9.945 \times 10^{-16} \, \text{J} \]

### Step 4: Calculate the kinetic energy of the electron

The kinetic energy \( KE \) of the electron can be calculated using the momentum we found earlier:

\[ KE = \frac{P_E^2}{2m} \]

Where:
- \( m = 9.1 \times 10^{-31} \, \text{kg} \) (mass of the electron)

Substituting the values:

\[ KE = \frac{(3.315 \times 10^{-25})^2}{2 \times (9.1 \times 10^{-31})} = 6.032 \times 10^{-20} \, \text{J} \]

### Summary of Results:
1. Momentum of the electron \( P_E = 3.315 \times 10^{-25} \, \text{kg m/s} \)
2. Momentum of the photon \( P_P = 3.315 \times 10^{-25} \, \text{kg m/s} \)
3. Energy of the photon \( E = 9.945 \times 10^{-16} \, \text{J} \)
4. Kinetic energy of the electron \( KE = 6.032 \times 10^{-20} \, \text{J} \)