The ratio of the wavelengths associated with electrons accelerated with the potentials of 64 V and 81 V, respectively is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 9:8 \)
An electron accelerated through a potential difference \( V \) has the de Broglie wavelength \( \lambda = h/\sqrt{2meV} \). For the same particle, \( \lambda \propto 1/\sqrt{V} \), so only the ratio of the potentials is needed.
\( \displaystyle \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{81}{64}} = \frac{9}{8} \)
So \( \lambda_1 : \lambda_2 = 9 : 8 \).