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If accelerating potential increases from20kv to 80kv in an electron microscope, its resolving power R would change to

Sayam Rahul , 7 Years ago
Grade 12th pass
anser 1 Answers
Arun

Last Activity: 7 Years ago

Dear Sayam
 
\lambda = h / P = (h / \small \sqrt2me)(1 / \small \sqrtv)
R.P. \small \alpha 1 / \lambda  \small \alpha  \small \sqrtv
R / R’ = \small \sqrt(20/ 80) = 1/ 2
R’ = 2R
hope it helps
Regards
Arun (askIITians forum expert)
 

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