Express the kinetic energy of an electron in terms of its wavelength using the de Broglie equation.

The Correct Answer and Explanation is:

To express the kinetic energy of an electron in terms of its wavelength, we can use the de Broglie equation and principles of classical mechanics.

Step 1: De Broglie Wavelength

The de Broglie equation relates the wavelength (λ\lambdaλ) of a particle to its momentum (ppp):λ=hp\lambda = \frac{h}{p}λ=ph

where:

hhh is Planck’s constant (h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}h=6.626×10−34J⋅s)
ppp is the momentum of the electron.

Momentum is defined as the product of the mass (mmm) of the electron and its velocity (vvv):p=mvp = mvp=mv

Thus, the de Broglie wavelength becomes:λ=hmv\lambda = \frac{h}{mv}λ=mvh

Step 2: Solving for Momentum

Rearranging this equation to solve for momentum:p=hλp = \frac{h}{\lambda}p=λh

Step 3: Kinetic Energy of an Electron

The classical kinetic energy (K.E.K.E.K.E.) of an electron is given by:K.E.=12mv2K.E. = \frac{1}{2}mv^2K.E.=21mv2

Using the relationship for momentum (p=mvp = mvp=mv), we can solve for vvv:v=pmv = \frac{p}{m}v=mp

Substituting this expression for velocity into the kinetic energy formula:K.E.=12m(pm)2=p22mK.E. = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}K.E.=21m(mp)2=2mp2

Step 4: Expressing Kinetic Energy in Terms of Wavelength

Now substitute the expression for momentum from the de Broglie equation (p=hλp = \frac{h}{\lambda}p=λh) into the equation for kinetic energy:K.E.=(hλ)22mK.E. = \frac{\left(\frac{h}{\lambda}\right)^2}{2m}K.E.=2m(λh)2

Simplifying:K.E.=h22mλ2K.E. = \frac{h^2}{2m\lambda^2}K.E.=2mλ2h2

Conclusion

Thus, the kinetic energy of an electron in terms of its de Broglie wavelength is:K.E.=h22mλ2K.E. = \frac{h^2}{2m\lambda^2}K.E.=2mλ2h2