Question

Electrons are accelerated through a p.d. of 150V. Given$m = 9.1 \times {10^{( - 31)}}Js$, the de Broglie wavelength associated with it is
A. $1.5A{}^\circ$
B. $1.0A{}^\circ$
C. $3.0A{}^\circ$
D. $0.5A{}^\circ$

Answer 1

We know that the de Broglie wavelength depends upon the mass of the particle as well as the velocity of the particle. Now, we need to use the law of conservation of energy and equate the kinetic energy of the particles with the energy produced from the potential difference and then apply it on the formula of de Broglie wavelength.

Complete step by step answer:
Here, we are provided that the potential difference through which the electrons are accelerated is 150V. The mass of the electrons is given as $9.1 \times {10^{( - 31)}}$kg. Now, the de-Broglie wavelength of a particle is given by the formula:
$\lambda = \dfrac{h}{{mv}}$
Here, $\lambda$ is the de-Broglie wavelength, $h$ is the Planck’s constant, $m$ is the mass of the particle and $v$ is the velocity of the particle. Now, the kinetic energy of the particles can be converted into potential energy and its formula is given as follows:
$\dfrac{1}{2}m{v^2} = eV$
Here, $m$ is the mass of the particle and $v$ is the velocity of the particle, $e$ is the electrical charge of the electron and $V$ is the potential difference. Thus;
$mv = \sqrt {2meV}$
Substituting the above value in the de-Broglie’s wavelength, we obtain;
$\lambda = \dfrac{h}{{\sqrt {2meV} }}$
We know that the value of the potential difference is 150V. Thus, the de-Broglie wavelength is as follows:

$$\lambda =\dfrac{6.6\times {{10}^{(-34)}}}{\sqrt{2\times 9.1\times {{10}^{(-31)}}\times 1.6\times {{10}^{(-19)}}\times 150}} \\\ \therefore \lambda =1A{}^\circ \\\$$

Hence option B is the correct answer.

Note: Here, due to the potential difference of the battery, the energy from the voltage is given to the electrons and gets converted into the kinetic energy of the electrons. If the surface we use is a metal surface, and the voltage difference is provided, then the ejection of electrons happens and the phenomenon is called photoelectric effect.