An electron microscope is operated at 40 kV. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The de Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda_e = \frac{12.27}{\sqrt{V}} \ \mathring{A} = \fra

Vedclass · Accepted Answer

$9.78 	imes 10^4$

Vedclass · Answer

(B) The de Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda_e = \frac{12.27}{\sqrt{V}} \ \mathring{A} = \frac{12.27}{\sqrt{40 	imes 10^3}} 	imes 10^{-10} \ m$.Calculating this,we get $\lambda_e \approx 6.13 	imes 10^{-12} \ m$.The resolving power $R$ of a microscope is inversely proportional to the wavelength $\lambda$ of the radiation used,i.e.,$R \propto \frac{1}{\lambda}$.Therefore,the ratio of the resolving power of the electron microscope $(R_1)$ to the optical microscope $(R_2)$ is $\frac{R_1}{R_2} = \frac{\lambda_2}{\lambda_1}$.Substituting the values: $\frac{R_1}{R_2} = \frac{6 	imes 10^{-7}}{6.13 	imes 10^{-12}} \approx 9.78 	imes 10^4$.

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

Similar Questions

The resolving power of a microscope depends upon:

The ratio of resolving powers of an optical microscope for two wavelengths $\lambda_1 = 4000\,\mathring{A}$ and $\lambda_2 = 6000\,\mathring{A}$ is

Assume that light of wavelength $600\, nm$ is coming from a star. The limit of resolution of a telescope whose objective has a diameter of $2\, m$ is $......... \times 10^{-7}\, rad$.

Wavelengths of light used in an optical instrument are $\lambda_1 = 4000 \; \mathring{A}$ and $\lambda_2 = 5000 \; \mathring{A}$. The ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$) is:

The resolving power of a compound microscope will be maximum when

Vedclass Test Series

Exam Paper Generator

Online Exam Module