An electron is accelerated from rest through a potential difference of $V = 120\,\text{V}$. Its de Broglie wavelength is closest to (Use the shortcut $\lambda\,[\text{\AA}] = \sqrt{150/V}$.)

1. An electron accelerated through potential $V$ gains kinetic energy $K = eV$. 2. Non-relativistic momentum: $p = \sqrt{2 m_e K} = \sqrt{2 m_e eV}$. 3. de Broglie wavelength: $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2 m_e eV}}$. Plugging the SI constants gives the standard NEET shortcut $\lambda\,[\text{\AA}] = \sqrt{150/V\,[\text{V}]}$. 4. Apply: $\lambda = \sqrt{150/120} = \sqrt{1.25} = 1.118\,\text{\AA} \approx 1.12\,\text{\AA}$. 5. Option A uses $V = 200$ in the shortcut. Option C uses $V = 75$. Option D drops the square root and reports $150/120 \times \text{something}$, a common arithmetic slip. _Source: NCERT Class 12 Physics Part 2, Ch 11, §11.8 (de Broglie hypothesis) + Example 11.3, p. 11–12._