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
$\text{Given: Electron accelerated through potentials } V_1 = 64~\text{V},~ V_2 = 81~\text{V}$
$\text{De Broglie wavelength: } \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}}$
$\text{Ratio of wavelengths: } \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{81}{64}} = \frac{9}{8}$
$\text{Answer: } \frac{\lambda_1}{\lambda_2} = \frac{9}{8}$