Question

The potential energy of particle of mass $m$ varies as
$ U(x) = \begin{cases} E_0 & \text{for $ 0 \le x \le 1 $ } \\[2ex] 0 & \text{for $ x > 1 $ } \end{cases} $
The de-Broglie wavelength of the particle in the range $ 0 \le x \le 1 $ is $ \lambda_1 $ and that in the range $ x > 1 $ is $ \lambda_2 $ . If the total energy of the particle is $ 2E_0 $ , find $ \frac{\lambda_{1}}{\lambda_{2}} $ .

• A. $ \sqrt{2} $
• B. $ \sqrt{3} $
• C. $ \sqrt{\frac{1}{2}} $
• D. $ \sqrt{\frac{2}{3}} $

Answer 1

Answer: (A)

We know that
$\lambda = \frac{h}{mv}$
$\lambda = \frac{h}{\sqrt{2mk}}$
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{k_2}{k_1}} \,\,$ ($\because h$ and $m$ are constants)
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{2k}{k}} = \sqrt{2}$