Question

We wish to see inside the atom. Assuming the atom to have a diameter of $100\text{pm}$, this means that one must be able to resolve a width of say $10\text{pm}$. If an electron-micro scope is used, the minimum electron energy required is about. (Assume the wavelength of light used in an electron microscope is nearly equal to the resolving power of the electron microscope.)

• A. $15\text{keV}$
• B. $1.5\text{keV}$
• C. $150\text{keV}$
• D. $1.5\text{MeV}$

Answer 1

The correct option is A $15\text{keV}$

The wavelength of light used in the electron microscope is nearly equal to the resolving power of the electron-microscope.

$\therefore \lambda =10\times {10}^{-12}\text{m}$

Hence, from the de-Broglie wavelength,

$\lambda =\frac{h}{p}=\frac{h}{mv}$

$\Rightarrow v=\frac{h}{\lambda m}=\frac{6.6\times {10}^{-34}}{1\times {10}^{-11}\times 9.1\times {10}^{-31}}$

$\Rightarrow v=7.25\times {10}^7{\text{ms}}^{-1}$

Minimum electron energy,

$KE=\frac{1}{2}m{v}^2=\frac{9.1\times {10}^{-31}\times (7.25\times {10}^7{)}^2}{2\times 1.6\times {10}^{-19}}=\frac{478.31\times {10}^{-17}}{3.2\times {10}^{-19}}$

$=149.47\times {10}^2\approx 150\times {10}^2$

$=15\text{keV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.