OnlineCBT
Chemistry · Structure of atom
हिन्दीCalculate the de Broglie wavelength of an alpha particle accelerated through a potential difference of \( 40 \nabla \text{ V} \) (mass of alpha particle \( = 6.64 \times 10^{-27} \text{ kg} \), charge \( q = 3.2 \times 10^{-19} \text{ C} \)).
View Answer & Solution
Correct answer · Option A
\( 0.143 \text{ \AA} \)
Explanation
Using \( \lambda = \frac{h}{\sqrt{2mqV}} \), substituting \( h = 6.626 \times 10^{-34} \), \( m = 6.64 \times 10^{-27} \), \( q = 3.2 \times 10^{-19} \), and \( V = 40 \), we obtain \( \lambda \approx 0.143 \text{ \AA} \).
