For the same objective, the ratio of least separation between two points to be distiguised by a microscope for light of `5000 Å` and electrons acclerated through 100V used as illuminating substance is __________ (neraly)

Options

A. `2`
B. `4xx10^(-2)`
C. `2xx10^(-6)`
D. `2xx10^(-4)`

Answer

D

Explanation

Resolving power `= (1)/(Delta x) = (2sin alpha)/(1.22lambda)`
`implies Delta x = (1.22lambda)/(2sin alpha)`
For `lambda = 5000 Å Delta x_(1) = (1.22xx5000xx10^(-10))/(25sin alpha)` …(1)
de-Brogle wavelength `lambda = (12.27)/(sqrt(v)) = (12.27)/(sqrt(100))`
`lambda = 1.227xx10^(-10)`
`Delta x_(2) = (1.22xx1.227xx10^(-10))/(2sin alpha)`
`(Delta x_(1))/(Delta x_(2)) = (1.227xx10^(-10))/(5000xx10^(-10)) = 2xx10^(-4)`