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 / \lambda (where \lambda is wavelength associated with electron wave) is given by :

Structure of Atom PYQ — JEE Main 2016 · Shift 0 · Q18 with Solution

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Correct answer: C

When an electron is accelerated through a potential difference $V$ , the kinetic energy gained is

\frac{1}{2}mv^2 = eV

Here, since the unit is mentioned as esu, we use the same relation in the given unit system.

The de Broglie relation is

\lambda = \frac{h}{p}

So,

\frac{h}{\lambda} = p

Thus we only need the momentum of the electron.

From

\frac{1}{2}mv^2 = eV

multiply both sides by $2m$ :

m^2v^2 = 2meV

But $mv = p$ , so

p^2 = 2meV

Hence,

p = \sqrt{2meV}

Therefore,

\frac{h}{\lambda} = \sqrt{2meV}

The correct option is:

\boxed{\sqrt{2meV}}

So, Option C is correct.