Question

If an electron initially at rest is subjected to a uniform electric field of \(4125\ \mathrm{NC^{-1}}\), then the de Broglie wavelength associated with the electron at time \(t=2\times10^{-9}\) s is \[ (\text{Planck's constant}=6.6\times10^{-34}\ \mathrm{Js}) \]

• A. \(12\ \text{\AA}\)
• B. \(2\ \text{\AA}\)
• C. \(10\ \text{\AA}\)
• D. \(5\ \text{\AA}\)

Hint:

For a charged particle initially at rest in a uniform electric field, \[ \boxed{ p=eEt. } \] Then use the de Broglie relation \[ \boxed{ \lambda=\frac{h}{p}. } \]

Answer 1

The Correct Option is D

Step 1: Find the momentum of the electron. The force acting on the electron is \[ F=eE. \] Hence, the momentum after time \(t\) is \[ p=Ft=eEt. \] Using \[ e=1.6\times10^{-19}\text{ C}, \] \[ E=4125\ \mathrm{NC^{-1}}, \] and \[ t=2\times10^{-9}\text{ s}, \] \[ p = 1.6\times10^{-19} \times4125 \times2\times10^{-9} = 1.32\times10^{-24}\ \mathrm{kg\,ms^{-1}}. \]

Step 2: Apply the de Broglie relation. The de Broglie wavelength is \[ \lambda=\frac{h}{p}. \] Thus, \[ \lambda = \frac{6.6\times10^{-34}} {1.32\times10^{-24}} = 5\times10^{-10}\text{ m}. \] Since \[ 1\ \text{\AA}=10^{-10}\text{ m}, \] \[ \lambda=5\ \text{\AA}. \]

Step 3: Write the answer. Hence, \[ \boxed{5\ \text{\AA}}. \] Thus, \[ \boxed{(D)} \] is the correct answer.