Question

Electrons used in an electron microscope are accelerated by a voltage of $25kV$ . If the voltage is increased to $100kV$ then the de-Broglie wavelength associated with the electrons would?
A. Increase by $2$ times
B. Decrease by $2$ times
C. Decrease by $4$ times
D. Increase by $4$ times

Answer 1

To solve this type of question, we will use the De Broglie wavelength in terms of accelerated potential and then compare the new and old wavelengths to get the required answer. If the charged particle is considered as the electron, then the de Broglie wavelength for the electron would be given as $\lambda = \dfrac{h}{{\sqrt {2meV} }}$ .

Formula used:
$E = e.V$
Where,
$E$ is the kinetic energy of the electron,
$e = 1.6 \times {10^{ - 19}}C$ is the charge of an electron and
$V$ is the accelerating potential applied.
$p = \sqrt {2mE}$
De-Broglie equation:
$\lambda = \dfrac{{12.27}}{{\sqrt V }}$
Where,
$V$ is the potential difference and
$\lambda$ is the wavelength associated with the particle.

Complete answer:
Initial and final wavelength are ${\lambda _1}$ ​and ${\lambda _2}$ ​having potential as ${V_1}$ ​and ${V_2}$ ​
Therefore,
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\dfrac{{{V_2}}}{{{V_1}}}}$

$$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\dfrac{{100}}{{25}}} \\\ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt 2 \\\ \Rightarrow {\lambda _2} = \dfrac{{{\lambda _1}}}{{\sqrt 2 }} \\\$$

So, from here we can say that the new wavelength has decreased by 2.
Hence, the correct option is B.

Note:
In order to solve this question, study the de Broglie hypothesis. The relationship between Kinetic Energy, potential difference and charge of the particle and also the relationship of kinetic energy with momentum is important.