Calculate Parameter Alpha in De Broglie Wavelength Variation

The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m e V}}$

where:

• $h$ is Planck's constant,
• $m$ is the mass of the electron,
• $e$ is the elementary charge,
• $V$ is the accelerating potential difference.

From this relation, it is clear that the de Broglie wavelength is inversely proportional to the square root of the potential difference: $\lambda \propto \frac{1}{\sqrt{V}}$

For the initial state with potential $V_1$ and wavelength $\lambda_1$: $\lambda_1 = \frac{h}{\sqrt{2m e V_1}} \quad \text{--- (1)}$

When the potential is changed to $V_2$, the new wavelength $\lambda_2$ increases by $50\%$ of $\lambda_1$: $\lambda_2 = \lambda_1 + 50\% \text{ of } \lambda_1 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1 = \frac{3}{2}\lambda_1$

For potential $V_2$, the wavelength equation is: $\lambda_2 = \frac{h}{\sqrt{2m e V_2}} \quad \text{--- (2)}$

Dividing equation (1) by equation (2), we get: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$

Substitute $\lambda_2 = \frac{3}{2}\lambda_1$ into the above equation: $\frac{\lambda_1}{\frac{3}{2}\lambda_1} = \sqrt{\frac{V_2}{V_1}}$ $\frac{2}{3} = \sqrt{\frac{V_2}{V_1}}$

Squaring both sides gives: $\frac{V_2}{V_1} = \frac{4}{9}$

Taking the reciprocal: $\frac{V_1}{V_2} = \frac{9}{4}$

Comparing this with the given condition $\frac{V_1}{V_2} = \frac{9}{\alpha}$, we find: $\alpha = 4$