A stream of electrons from a heat filament was passed between two charge plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of $$\frac{h}{\lambda}$$ (where $$\lambda$$ is wavelength associated with the electron wave) is given by:

First, recall the de-Broglie relation, which connects the momentum $$p$$ of any particle with its wavelength $$\lambda$$ by the formula

$$\lambda \;=\;\frac{h}{p}\;.$$

From this same relation, it follows directly that

$$\frac{h}{\lambda}\;=\;p\;.$$

Therefore, in order to find $$\dfrac{h}{\lambda}$$ for the electron beam, we only need to calculate the magnitude of the momentum $$p$$ that each electron acquires after being accelerated through the given potential difference $$V$$.

When an electron of charge $$e$$ is accelerated from rest through an electric potential difference $$V$$ (expressed here in electrostatic units, i.e. esu), the work done by the electric field becomes its kinetic energy. Hence we can write

$$eV \;=\; \text{kinetic energy of the electron}.$$

For a particle of mass $$m$$ moving with speed $$v$$, the classical kinetic-energy formula is

$$\text{K.E.}\;=\;\frac{1}{2}\,m v^{2}.$$

Equating the electrical work to the kinetic energy, we have

$$eV \;=\;\frac{1}{2}\,m v^{2}\;.$$

Now we solve step-by-step for the velocity $$v$$. Multiplying both sides by $$2$$ gives

$$2\,eV \;=\;m v^{2}\;.$$

Dividing both sides by $$m$$, we obtain

$$v^{2} \;=\;\frac{2\,eV}{m}\;.$$

Taking the positive square root (since speed is positive), we get

$$v \;=\;\sqrt{\frac{2\,eV}{m}}\;.$$

The linear momentum $$p$$ of the electron is given by the basic definition

$$p \;=\;m v\;.$$

Substituting the expression for $$v$$ that we have just derived, we find

$$p \;=\;m \,\sqrt{\frac{2\,eV}{m}} \;=\;\sqrt{m^{2}}\,\sqrt{\frac{2\,eV}{m}} \;=\;\sqrt{m}\,\sqrt{\frac{2\,eV\,m}{m}} \;=\;\sqrt{2\,m\,eV}\;.$$

(In the third step above, the mass $$m$$ inside and outside the square roots was combined so that all factors lie neatly under a single radical.) Thus the magnitude of the momentum acquired by each electron is

$$p \;=\;\sqrt{2\,m\,eV}\;.$$

But earlier we noted that $$\dfrac{h}{\lambda}=p$$. Therefore, substituting the value of $$p$$ obtained here, we finally have

$$\frac{h}{\lambda}\;=\;\sqrt{2\,m\,eV}\;.$$

Among the four listed options, the expression $$\sqrt{2\,meV}$$ matches exactly with the result we have derived.

Hence, the correct answer is Option B.