Question

The work done to accelerate an electron from rest so that it can have a de Broglie wavelength of 6600 Å is nearly (Planck's constant = \(6.6 \times 10^{-34}\) Js and mass of electron = \(9 \times 10^{-31}\)kg)

• A. \(5.56 \times 10^{-25}\) eV
• B. 1.88 eV
• C. \(5.56 \times 10^{-25}\) J
• D. 1.88 J

Hint:

There are two key formulas for kinetic energy. The classical one is \(K.E. = \frac{1}{2}mv^2\). The one in terms of momentum, \(K.E. = p^2/2m\), is often more useful in quantum physics problems where the de Broglie wavelength (which gives momentum) is known.

Answer 1

The Correct Option is C

Step 1: Relate work done to kinetic energy.
The work done to accelerate an electron from rest is equal to the final kinetic energy (K.E.) of the electron, according to the work-energy theorem.
\( W = K.E. = \frac{1}{2}mv^2 \), where m is mass and v is the final velocity.
Step 2: Find the momentum of the electron from its de Broglie wavelength.
The de Broglie wavelength (\(\lambda\)) is related to momentum (p) by the equation \( \lambda = \frac{h}{p} \), where h is Planck's constant.
We are given \( \lambda = 6600 \text{ Å} = 6600 \times 10^{-10} \text{ m} = 6.6 \times 10^{-7} \text{ m} \).
We can calculate the momentum: \( p = \frac{h}{\lambda} = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-7}} = 1 \times 10^{-27} \) kg·m/s.
Step 3: Relate kinetic energy to momentum.
The kinetic energy can be expressed in terms of momentum as \( K.E. = \frac{p^2}{2m} \).
We have the momentum p and the mass of the electron \( m = 9 \times 10^{-31} \) kg.
Substitute these values to find the kinetic energy.
\( K.E. = \frac{(1 \times 10^{-27})^2}{2 \times (9 \times 10^{-31})} = \frac{1 \times 10^{-54}}{18 \times 10^{-31}} \).
\( K.E. = \frac{1}{18} \times 10^{-54 - (-31)} = \frac{1}{18} \times 10^{-23} \) J.
\( \frac{1}{18} \approx 0.0555... \)
So, \( K.E. \approx 0.0556 \times 10^{-23} = 5.56 \times 10^{-25} \) J.
Since Work Done = K.E., the work done is approximately \( 5.56 \times 10^{-25} \) J.