CBSE 2026 · Region 1 · Set 3 · Q20 · 2 marks
In an electron microscope, accelerated electrons have wavelength of $\displaystyle 0.011$ nm. Calculate the voltage through which electrons were accelerated to attain this wavelength. (Take $\displaystyle \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ )
Marking-scheme solution
de-Broglie wavelength $\displaystyle \lambda = \dfrac{h}{\sqrt{2meV}}$$\displaystyle V = \dfrac{h^2}{2me \times \lambda^2}$$\displaystyle V = \dfrac{(6.6 \times 10^{-34})^2}{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} (0.011 \times 10^{-9})^2}$$\displaystyle V = 12500\ \text{Volts}$
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.