Calculate the de Broglie wavelength associated with an electron moving with a speed $v = 5.4\times 10^{6}\,\text{m\,s}^{-1}$. (Take $h = 6.63\times 10^{-34}\,\text{J\,s}$, electron mass $m_e = 9.11\times 10^{-31}\,\text{kg}$.)

1. The de Broglie wavelength of a particle is $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$. 2. Compute the momentum: $p = m_e v = (9.11\times 10^{-31})(5.4\times 10^{6}) = 4.92\times 10^{-24}\,\text{kg\,m\,s}^{-1}$. 3. Then $\lambda = \dfrac{6.63\times 10^{-34}}{4.92\times 10^{-24}} = 1.35\times 10^{-10}\,\text{m}$. 4. Convert to nanometres: $1.35\times 10^{-10}\,\text{m} = 0.135\,\text{nm}$, which rounds to $0.13\,\text{nm}$. 5. Distractors come from common slips: option A drops a factor of 2 in $p$; option C is the correct value off by a factor of 10 (decimal slip); option D uses $1.35\times 10^{-9}$ by misreading the exponent. None apply here. _Source: NCERT Class 12 Physics Part 2, Ch 11, Example 11.3 (part a), p. 12 (solution on p. 12 §11.8)._