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Step-by-Step Solution
Step 1: Identify the Relevant Concept
The resolution of an electron microscope depends on the wavelength of the electrons. The de Broglie wavelength $ \lambda $ of an electron is given by:
$ \lambda = \frac{h}{p} $,
where $h$ is Planck’s constant and $p$ is the momentum of the electron.
Step 2: Express Momentum in Terms of Kinetic Energy
The kinetic energy $KE$ of an electron can be written in terms of its momentum $p$ as:
$ KE = \frac{p^2}{2m} $,
where $m$ is the mass of the electron. From this, the momentum can be expressed as:
$ p = \sqrt{2m \, KE} $.
Step 3: Relate Wavelength to Kinetic Energy
Combining $ \lambda = \frac{h}{p} $ with $p = \sqrt{2m \, KE}$, we get:
$ \lambda = \frac{h}{\sqrt{2m \, KE}} $.
Rearranging for $KE$ gives:
$ KE = \frac{h^2}{2m \, \lambda^2} $.
Step 4: Substitute Numerical Values
Planck’s constant, $h = 6.6 \times 10^{-34}\,\text{J·s}$
Mass of electron, $m = 9.1 \times 10^{-31}\,\text{kg}$
Desired wavelength, $\lambda = 7.5 \times 10^{-12}\,\text{m}$
Thus,
$ KE = \frac{(6.6 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (7.5 \times 10^{-12})^2} \,\text{J}$.
Step 5: Perform the Calculation and Convert to Electronvolts
Calculate the energy in joules and then convert from joules to electronvolts (eV) using the relation $1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$. Finally, convert eV to keV by dividing by $10^3$.
The approximate result comes out to about $25\,\text{keV}$.
Final Answer
Hence, the minimum electron energy required for a resolution of about $7.5 \times 10^{-12}\,\text{m}$ is approximately 25 keV.
