It is desired to obtain a diffraction pattern for electrons using a diffraction grating with lines separated by 10 nm. The mass of an electron is `9.11 xx 10^(-31)kg.`
What is the approximate kinetic energy of electrons that would be diffracted by such a grating?

Options

A. `1.5xx10^(-6)eV`
B. `1.5xx10^(-2)eV`
C. `1.5xx10^(-4)eV`
D. `1.5xx10^(2)eV`

Answer

To solve the problem of determining the approximate kinetic energy of electrons that would be diffracted by a grating with lines separated by 10 nm, we can follow these steps:

### Step 1: Understand the relationship between wavelength and diffraction
For diffraction to occur, the wavelength of the electrons must be comparable to the spacing of the grating. In this case, the grating spacing is given as 10 nm, which we will consider as the wavelength (λ) of the electrons.

### Step 2: Use the de Broglie wavelength formula
The de Broglie wavelength (λ) of a particle is given by the equation:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 3: Relate momentum to kinetic energy
The momentum \( p \) of an electron can be expressed in terms of its kinetic energy \( E \) using the relation:
\[ p = \sqrt{2mE} \]
where \( m \) is the mass of the electron.

### Step 4: Substitute momentum into the de Broglie equation
Substituting the expression for momentum into the de Broglie equation gives:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

### Step 5: Rearrange to find kinetic energy
Squaring both sides and rearranging for kinetic energy \( E \):
\[ \lambda^2 = \frac{h^2}{2mE} \implies E = \frac{h^2}{2m\lambda^2} \]

### Step 6: Plug in the known values
Now we can substitute the known values:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \)
- Wavelength \( \lambda = 10 \, \text{nm} = 10 \times 10^{-9} \, \text{m} \)

Calculating \( \lambda^2 \):
\[ \lambda^2 = (10 \times 10^{-9})^2 = 10^2 \times 10^{-18} = 10^{-16} \, \text{m}^2 \]

Now substituting into the energy equation:
\[ E = \frac{(6.63 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times (10^{-16})} \]

Calculating \( h^2 \):
\[ h^2 = (6.63 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{Js}^2 \]

Now substituting this value into the equation:
\[ E = \frac{4.39 \times 10^{-67}}{2 \times 9.11 \times 10^{-31} \times 10^{-16}} \]

Calculating the denominator:
\[ 2 \times 9.11 \times 10^{-31} \times 10^{-16} = 1.822 \times 10^{-46} \]

Now calculating \( E \):
\[ E = \frac{4.39 \times 10^{-67}}{1.822 \times 10^{-46}} \approx 2.41 \times 10^{-21} \, \text{J} \]

### Step 7: Convert Joules to electron volts
To convert energy from Joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \, \text{(in eV)} = \frac{2.41 \times 10^{-21}}{1.6 \times 10^{-19}} \approx 0.0150 \, \text{eV} = 1.5 \times 10^{-2} \, \text{eV} \]

### Conclusion
Thus, the approximate kinetic energy of electrons that would be diffracted by such a grating is:
\[ \boxed{1.5 \times 10^{-2} \, \text{eV}} \]