An electron is accelerated from rest through a potential difference \(V\). Its de Broglie wavelength then becomes 0.11 nm. The value of \(V\) is (Take \(h = 6.6\times10^{-34}\ \mathrm{J\,s}\), \(e = 1.6\times10^{-19}\ \mathrm{C}\) and mass of the electron \(= 9.0\times10^{-31}\ \mathrm{kg}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 125 V
The de Broglie relation gives the momentum of the electron as \( p=h/\lambda \). Its kinetic energy is \( K=p^{2}/2m \). An electron accelerated through a potential difference \( V \) gains kinetic energy \( K=eV \), so \( V=K/e \).
\( \lambda=0.11\ \mathrm{nm}=1.1\times10^{-10}\ \mathrm{m} \)
\( \begin{aligned} p &= \frac{h}{\lambda} \\ &= \frac{6.6\times10^{-34}}{1.1\times10^{-10}} \\ &= 6.0\times10^{-24}\ \mathrm{kg\,m\,s^{-1}} \end{aligned} \)
\( \begin{aligned} K &= \frac{p^{2}}{2m} \\ &= \frac{(6.0\times10^{-24})^{2}}{2\times9.0\times10^{-31}} \\ &= 2.0\times10^{-17}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} V &= \frac{K}{e} \\ &= \frac{2.0\times10^{-17}}{1.6\times10^{-19}} \\ &= 125\ \mathrm{V} \end{aligned} \)
From the above, \( V=125\ \mathrm{V} \). Hence, a small voltage gives an electron a wavelength of about the size of an atom.
NCERT (2026-27):
Electron Emission – “… One electron volt is the energy gained by an electron when it has been accelerated by a potential difference of 1 volt, so that 1 eV = 1.602 ×10–19 J.”
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as”
\( \displaystyle \lambda=\frac{h}{p}=\frac{h}{mv} \)\( \qquad (11.5) \)
“where m is the mass of the particle and v its speed.”