Question

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by:
A: $2meV$
B: $\sqrt {meV}$
C: $\sqrt {2meV}$
D: $meV$

Answer 1

As you can clearly see in the options that energy term (i.e. eV) is mentioned thus, it indicates that we will have to find out the relationship between $\dfrac{h}{\lambda }$and energy where h is the Planck’s constant and $\lambda$is the wavelength of electron wave.

Complete Step by step answer:
De Broglie wave equation and kinetic energy equation can be used to derive the relation between $\dfrac{h}{\lambda }$ and energy. de Broglie equation simply states that a matter can also act as a wave like light or radiation which also behave like waves and particles. This equation can also be used to further explain that a beam of electrons also diffract just as a beam of light. The de Broglie equation can be written as follows:
$\lambda = \dfrac{h}{{mv}} = \dfrac{h}{p}$
(h = Planck’s constant ($6.626 \times {10^{ - 34}}{m^2}kg/s$), $\lambda$=wavelength in m, m = mass of particle in kg, v = velocity of particle in ms-1, p = momentum)
Now, let us recall the kinetic energy equation. The kinetic energy (KE) of any object refers to the energy which it possesses owing to its motion. The kinetic energy equation can be written as follows:
$K.E = \dfrac{1}{2}m{v^2}$
K.E can also be written as eV (electron volt). So,

$$eV = \dfrac{1}{2}m{v^2} \\\ \Rightarrow mv = p \\\ \therefore eV = \dfrac{{{p^2}}}{{2m}} \\\$$

Now comparing kinetic energy and de Broglie equations:
From de Broglie equation:
$\dfrac{h}{\lambda } = p$ ......................................(1)
From Kinetic Energy equation:
$p = \sqrt {2meV}$ ……………………………………………..(2)
Now, from equations (1) and (2)
$\dfrac{h}{\lambda } = \sqrt {2meV}$

Thus, the correct answer is Option C.

Note: de Broglie equation is proved to be good regarding microscopic particles like electrons, protons and neutrons but it fails to work regarding large size objects as they possess more mass and thus their wavelength become so smaller that is not easy to measure.