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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2019
- Medium
- Numerical
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 \(7.5 \times 10^{-12}\, m\), the minimum electron energy required is close to .............. \(keV\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
\(\mathrm{k}=\frac{\mathrm{P}^{2}}{2 \mathrm{m}}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} \lambda^{2}}\)
\( = \frac{{{{\left( {6.62 \times {{10}^{ - 34}}} \right)}^2}}}{{2 \times 9.1 \times {{10}^{ - 31}} \times {{\left( {7.5 \times {{10}^{ - 12}}} \right)}^2}}}\) \( \times \frac{1}{{1.6 \times {{10}^{ - 19}}}}{\text{eV}}\)
Alternate method
\(\lambda=\sqrt{\frac{150}{V}}\)
\(\Rightarrow \quad 7.5 \times 10^{-12}=\sqrt{\frac{150}{V}}\)
\(V=\frac{80}{3} \mathrm{kV}\)
Energy \(=\frac{80}{3} \mathrm{keV}=25 \mathrm{keV}\)
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