Question

An electron at rest is accelerated through 10 kV potential. The de Broglie wavelength (in A) of the electron is .............

Hint:

The de Broglie wavelength for particles like electrons can be calculated using their momentum and Planck's constant. Remember to use the correct units and convert them if necessary.

Answer 1

The de Broglie wavelength (\( \lambda \)) of a particle is given by the de Broglie relation:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck’s constant and \( p \) is the momentum of the particle.

Step 1: Calculate the momentum of the electron.
The kinetic energy \( K.E. \) of the electron after being accelerated by a potential \( V \) is:
\[ K.E. = eV \]
where \( e \) is the charge of the electron and \( V \) is the potential.
Given \( V = 10 \, \text{kV} = 10^4 \, \text{V} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), the kinetic energy is:
\[ K.E. = (1.6 \times 10^{-19})(10^4) = 1.6 \times 10^{-15} \, \text{J} \]
Since the electron is initially at rest, all the kinetic energy is converted into momentum.
The relation between kinetic energy and momentum is:
\[ K.E. = \frac{p^2}{2m} \]
where \( m = 9.11 \times 10^{-31} \, \text{kg} \).

Solving for \( p \):
\[ p = \sqrt{2mK.E.} \]
\[ p = \sqrt{2(9.11 \times 10^{-31})(1.6 \times 10^{-15})} \]

So, \( p = 5.40 \times 10^{-23} \) kg·m/s

Step 2: Calculate the de Broglie wavelength.
Using the relation:
\[ \lambda = \frac{h}{p} \]

Substituting \( h = 6.63 \times 10^{-34} \) J·s and \( p = 5.40 \times 10^{-23} \) kg·m/s:

\[ \lambda = \frac{6.63 \times 10^{-34}}{5.40 \times 10^{-23}} \]
\[ \lambda = 1.23 \times 10^{-11} \, \text{m} \]
Convert to angstroms (\( 1 \, \text{Å} = 10^{-10} \, \text{m} \)):
\[ \lambda = 1.23 \times 10^{-11} \times 10^{10} = 0.123 \, \text{Å} \]

Final Answer:
\[ \boxed{0.123 \, \text{Å}} \]