Question 189a2

Jan 6, 2017

$\textsf{{\lambda}_{2} = {\lambda}_{1} / 2}$

The wavelength is halved.

Explanation:

The de Broglie expression gives us:

$\textsf{{\lambda}_{1} = \frac{h}{m {v}_{1}}} \text{ } \textcolor{red}{\left(1\right)}$

When the electron of mass m and charge e is accelerated through a potential difference of $\textsf{{V}_{1}}$ volts we get:

$\textsf{e {V}_{1} = \frac{1}{2} m {v}_{1}^{2}} \text{ } \textcolor{red}{\left(2\right)}$

Where $\textsf{{v}_{1}}$ is the velocity of the electron.

The accelerating voltage is now increased by a factor of 4.

From sf(color(red)((2)) this means that the kinetic energy of the electron must also be increased by a factor of 4.

Since the kinetic energy of the electron depends on $\textsf{{v}_{1}^{2}}$ this means that the velocity of the electron has increased by a factor of 2. (2 x 2 = 4).

From $\textsf{\textcolor{red}{\left(1\right)}}$ you can see that doubling the velocity must, therefore, decrease the wavelength $\textsf{\lambda}$ by a factor of 2.

A more formal proof is given below:

From $\textsf{\textcolor{red}{\left(2\right)} \Rightarrow}$

$\therefore$$\textsf{{v}_{1}^{2} = \frac{2 e {V}_{1}}{m}}$

$\therefore$$\textsf{{v}_{1} = \sqrt{\frac{2 e {V}_{1}}{m}}}$

Subs. into $\textsf{\textcolor{red}{\left(1\right)} \Rightarrow}$

sf(lambda_1=(h)/(m.sqrt((2eV_1)/(m))#

The voltage is now increased by a factor of 4:

$\textsf{{V}_{2} = 4 {V}_{1}}$

$\therefore$$\textsf{{\lambda}_{2} = \frac{h}{m . \frac{\sqrt{2 e .4 {V}_{1}}}{m}}}$

$\textsf{{\lambda}_{1} / {\lambda}_{2} = \frac{\cancel{h}}{\cancel{m} . \sqrt{\frac{2 e {V}_{1}}{m}}} \times \cancel{m} . \frac{\sqrt{\frac{8 e {V}_{1}}{m}}}{\cancel{h}}}$

$\textsf{{\lambda}_{1} / {\lambda}_{2} = \sqrt{\frac{\cancel{8} \cancel{e} \cancel{{V}_{1}}}{\cancel{m}} \times \frac{\cancel{m}}{\cancel{2} \cancel{e} \cancel{{V}_{1}}}}}$

$\textsf{{\lambda}_{1} / {\lambda}_{2} = \sqrt{4} = 2}$

$\therefore$$\textsf{{\lambda}_{2} = {\lambda}_{1} / 2}$

This seems a reasonable result, as increasing the voltage should increase the kinetic energy and hence, the velocity of the electron.

The de Broglie expression tells us that this must decrease the wavelength.