Question

For particles having same K.E., the de-Broglie wavelength is

• A. Directly proportional to its velocity
• B. Inversely proportional to its velocity
• C. Independent of its mass and velocity
• D. Unpredictable

Answer 1

Answer: (A)

Directly proportional to its velocity

Answer 2

$\frac{\lambda_{1}}{\lambda_{2}} = \frac{m_{2}v_{2}}{m_{1}v_{1}}$ $\frac{\frac{1}{2}m_{2}v_{2}^{2}}{\frac{1}{2}m_{1}v_{1}^{2}}\cdot\frac{v_{1}}{v_{2}}$ $= \frac{K.E_{1}}{K.E_{2}} = \frac{v_{1}}{v_{2}}$ As $K.E_{1} = K.E_{2}$ $\therefore\frac{\lambda_{1}}{\lambda_{2}} = \frac{v_{1}}{v_{2}}$ or $\lambda\propto v$