Ncert Exercises & Solutions

Question 11.33: An electron microscope uses electrons accelerated by a voltage of 50 W. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Hint: $\lambda=\frac{h}{\sqrt{2mE}}$
Step 1: Find the kinetic energy of the electron.

The kinetic energy of the electron is given as:
E = eV = 1.6 x 10^-19 x 50 x 10³ = 8 x 10^-15 J

Step 2: Find the De Broglie wavelength of an electron.
De Broglie wavelength is given by the relation:

$\begin{aligned} λ & = \frac{h}{\sqrt{2 mE}} \\ = \frac{6 .6 \times 10^{- 34}}{\sqrt{2 \times 9 .11 \times 10^{- 31} \times 8 \times 10^{- 15}}} \\ = 5 .467 \times 10^{- 12} m \end{aligned}$

Step 3: Compare the resolving power of a microscope.
This wavelength is nearly 10⁵ times less than the wavelength of yellow light.
The resolving power of a microscope is inversely proportional to the wavelength of light used.
Thus, the resolving power of an electron microscope is nearly 10⁵ times that of an optical microscope.

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