CBSE 2022 · Region 1 · Set 3 · Q7 · 3 marks
The de-Broglie wavelength associated with an electron is $\displaystyle 0.30$ nm. Calculate : (a) the speed, and (b) the kinetic energy (in eV), of the electron.
Marking-scheme solution
(a)
b)a) $\displaystyle \lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
$$\mathrm{v}=\frac{\mathrm{h}}{\mathrm{~m} \lambda}=\frac{$\displaystyle 6.63$ \times $\displaystyle 10$^{-$\displaystyle 34$}}{$\displaystyle 9.1$ \times $\displaystyle 10$^{-$\displaystyle 31$} \times $\displaystyle 0.30$ \times $\displaystyle 10$^{-$\displaystyle 9$}}
$$=$\displaystyle 2.41$ \times $\displaystyle 10$^{$\displaystyle 6$} \mathrm{~m} / \mathrm{s}
$$b) K.E. $\displaystyle =\frac{1}{2} m v^{2}$
$$=\frac{1}{2} \times $\displaystyle 9.1$ \times $\displaystyle 10$^{-$\displaystyle 31$} \times\left($\displaystyle 2.41$ \times $\displaystyle 10$^{$\displaystyle 6$}\right)^{$\displaystyle 2$} \mathrm{~J}
$$=$\displaystyle 26.42$ \times $\displaystyle 10$^{-$\displaystyle 19$} \mathrm{~J}
$$K.E in eV
$$=\frac{$\displaystyle 26.42$ \times $\displaystyle 10$^{-$\displaystyle 19$}}{$\displaystyle 1.6$ \times $\displaystyle 10$^{$\displaystyle 19$}}=$\displaystyle 16.5$ \mathrm{eV}
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.