In an electron microscope if the potential is increased from $20$ kV to $80$ kV, the resolving power R of the microscope will become :

\frac{R}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 R

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2 R$
(b) Resolving power $\propto \frac{1}{λ}$ and $\propto \sqrt{v}$
$∴ R . P \propto \sqrt{v}$
$\Rightarrow (R . P)_{2} = (R . P)_{1} \times \sqrt{\frac{80 k V}{20 k V}}$
$= 2 (R . P)_{1}$ .

Suggest Corrections

Similar questions

Q. In an electron microscope the accelerating voltage is increased from 20 kV to 80 kV, the resolving power ofthe microscope will become.

Q. If accelerating potential increases from

20 K V

80 K V

in an electron microscope, its resolving power

R

would change to

Q. In an electron microscope the accelerating voltage is increased from 20 kV to 80 kV, the resolving power of the microscope will change from R to

An electron microscope works on principle of diffraction of electron beam .In the electron microscope if the Potential is increased from 20 kV to 80 kV, the resolving power R of the microscope will become

Q. The resolving power of an electron microscope operated at 16 kV is R. The resolving power of the electron microscope when operated at 4 kV is

Join BYJU'S Learning Program

In an electron microscope if the potential is increased from 20kV to 80kV, the resolving power R of the microscope will become :

The correct option is A 2R(b) Resolving power ∝1λ and ∝√v∴R.P∝√v⇒(R.P)2=(R.P)1×√80kV20kV=2(R.P)1.

In an electron microscope if the potential is increased from $20$ kV to $80$ kV, the resolving power R of the microscope will become :

The correct option is A $2 R$
(b) Resolving power $\propto \frac{1}{λ}$ and $\propto \sqrt{v}$
$∴ R . P \propto \sqrt{v}$
$\Rightarrow (R . P)_{2} = (R . P)_{1} \times \sqrt{\frac{80 k V}{20 k V}}$
$= 2 (R . P)_{1}$ .