If accelerating potential increases from $20 K V$ to $80 K V$ in an electron microscope, its resolving power $R$ would change to

\frac{R}{4}

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4 R

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2 R

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\frac{R}{2}

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Solution

The correct option is C $2 R$
$\frac{1}{2} m v^{2} = e V$
$m v = \sqrt{2 e V m}$
And $λ = \frac{h}{m V}$
$\frac{λ_{0}}{λ_{1}} = \frac{\sqrt{2 e V_{1} m}}{\sqrt{e V_{2} m}}$
$\frac{λ_{2}}{λ_{1}} = \frac{1}{2}$
$∴ λ_{2} = \frac{λ_{1}}{2}$
$R \propto \frac{1}{λ}$
so $R$ would change to $2 R$ .

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

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