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

To find the work done to accelerate an electron such that it has a de Broglie wavelength of 6600 Å, we can use the relationship between wavelength, momentum, and kinetic energy.

1. The de Broglie wavelength \(\lambda\) is given by the formula: 
   \(\lambda = \frac{h}{p}\) 
   where \(h = 6.6 \times 10^{-34}\) Js is Planck's constant and \(p\) is the momentum.
2. The momentum \(p\) of the electron can be expressed as: 
   \(p = \frac{h}{\lambda}\) 
   Substituting the given wavelength \(\lambda = 6600 \times 10^{-10}\) m into the equation: 
   \(p = \frac{6.6 \times 10^{-34}}{6600 \times 10^{-10}}\)
3. Simplifying the momentum: 
   \(p = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-7}} = 10^{-27} \text{kg m/s}\)
4. The kinetic energy \(E_k\) of an electron can be given by: 
   \(E_k = \frac{p^2}{2m}\) 
   where \(m = 9 \times 10^{-31}\) kg is the mass of the electron.
5. Substituting the values in the kinetic energy equation: 
   \(E_k = \frac{{(10^{-27})}^2}{2 \times 9 \times 10^{-31}}\)
6. Calculating \(E_k\): 
   \(E_k = \frac{10^{-54}}{1.8 \times 10^{-30}} = \frac{1}{1.8} \times 10^{-24} \approx 5.56 \times 10^{-25} \text{J}\)

Hence, the work done to accelerate the electron is nearly \(5.56 \times 10^{-25}\) J, which matches option \(5.56 \times 10^{-25}\) J.