Question

An electron microscope is operated at $40Kv$ . The ratio of resolving power of this microscope and another one which uses yellow light of wavelength $6\times {10}^7$ is :

• A. $7\times {10}^{-25}$
• B. $9.78\times {10}^4$
• C. $9.78\times {10}^{-4}$
• D. $9.78\times {10}^{-6}$

Answer 1

The correct option is A $7\times {10}^{-25}$

$\begin{array}{c}Thereslovingpowerofamicrospcopeisinverselyproportinaltothewavelengthoflightused,\\ forfirstelectronmicroscope,\\ \lambda =\frac{h}{\sqrt{2m_{e}E}}\\ \lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.11\times {10}^{-31}\times 1.6\times {10}^{-19}\times 40\times {10}^3}}=3.5\times {10}^{-10}m\\ for\sec ondmicroscop,\\ {\lambda }^{\prime }=\frac{c}{v}=\frac{3\times {10}^8}{6\times {10}^{-7}}=0.5\times {10}^{15}m\\ \frac{\lambda }{{\lambda }^{\prime }}=\frac{3.5\times {10}^{-10}}{0.5\times {10}^{15}}=7\times {10}^{-25}\end{array}$

Hence,

option $\left(A\right)$ is correct answer.