1.

If accelerating potential increases from 20 kV to 80 kV in an electron microscope, its resolving power R would change to :

Answer»

`( R)/(4)`
4R
2R
`(R )/(2)`

Solution :(c ) `R prop (a)/(lambda) prop (1)/(lambda)`
`THEREFORE lambda=(h)/(sqrt(2m_(e)V))prop(1)/(sqrt(V))` `(R.)/(R ) = sqrt((80KV)/(20KV))= 2`
`therefore R. = 2R`


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