Question

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5\times {10}^{-12}\mathrm{m}$, the minimum electron energy required is close to :

• A. $1\mathrm{keV}$
• B. $25\mathrm{keV}$
• C. $500\mathrm{keV}$
• D. $100\mathrm{keV}$

Answer 1

The correct option is B

$25\mathrm{keV}$

Explanation for correct option:

Option (b) $25\mathrm{keV}$

Step 1: Given data

• Given wavelength $=7.5\times {10}^{-12}\mathrm{m}$
• Mass of electron $=9.1\times {10}^{-31}\mathrm{Kg}$

Step 2: Formula used

• By using the De Broglie equation:

$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{P}}$

• $\Rightarrow \mathrm{P}=\frac{\mathrm{h}}{\mathrm{\lambda }}$
• Now, to find Kinetic energy, we use the formula given below:
• $\mathrm{KE}=\frac{{\mathrm{P}}^2}{2\mathrm{m}}=\frac{(\mathrm{h}/\mathrm{\lambda }{)}^2}{2\mathrm{m}}$

Step 3: Find Kinetic Energy

• Now, we will substitute the given values in the formula given below:

$\mathrm{KE}=\frac{(\frac{\mathrm{h}}{\mathrm{\lambda }}{)}^2}{2\mathrm{m}}$

$\mathrm{K}.\mathrm{E}.=\frac{\left\{\frac{6.6\times {10}^{-34}}{7.5\times {10}^{-12}}\right\}}{2\times 9.1\times {10}^{-31}}$

$\mathrm{K}.\mathrm{E}.=25\mathrm{keV}$

Thus, the minimum electron energy required is $25\mathrm{keV}$