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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 .............. \(keV\)

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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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