Question

In an electron microscope, electrons accelerated by a voltage of $50\text{kV}$ are used. In an optical microscope of the same aperture, yellow light of wavelength $5900\mathring{\text{A}}$ is used. The resolving power of the electron microscope will be $n$ times higher than that of the optical microscope, where $n$ is of the order of

• A. ${10}^2$
• B. ${10}^3$
• C. ${10}^4$
• D. ${10}^5$

Answer 1

The correct option is D ${10}^5$

Given, $V=50\text{kV};{\lambda }_y=5900\mathring{A}=5.9\times {10}^{-7}\text{m}$

Now, the wavelength of the light used in the electron-microscope is,

${\lambda }_e=\frac{h}{\sqrt{2meV}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times (9\times {10}^{-31})\times (1.6\times {10}^{-19})\times (50\times {10}^3)}}$

$=\frac{6.6\times {10}^{-34}}{12.06\times {10}^{-23}}\approx 0.55\times {10}^{-11}\text{m}\text{(or)}$

$\approx 5.5\times {10}^{-12}\text{m}$

$\because {\lambda }_y=590\text{nm}=5.9\times {10}^{-7}\text{m}$

$\because \text{The resolving power(R.P) of a microscope}\propto \frac{1}{\lambda }$

$\therefore \frac{\text{R.P. of electron microscope}}{\text{R.P. of optical microscope}}=\frac{{\lambda }_y}{{\lambda }_e}$

$=\frac{5.9\times {10}^{-7}}{5.5\times {10}^{-12}}={10}^5$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.

Why this question? Trick: To compare the resolving power, always remember that it is inversely proportional to the wavelength of light used. Note: As it’s evident you have noticed that electron microscopes have significantly larger resolving power.