An electron of mass $$m$$ is moving in an electric field $$\vec{E} = -2E_o \hat{i}$$ ($$E_o$$ = constant > 0), with an initial velocity $$\vec{V} = v_o \hat{i}$$ ($$v_o$$ = constant > 0). If $$\lambda_o = \frac{h}{4mv_o}$$, its de Broglie wavelength at time $$t$$ is _______. ($$e$$ = charge of electron)

Solution :

Electric field :

$$\vec E = -2E_0\hat i$$

Force on electron :

$$\vec F = q\vec E$$

Since charge of electron is :

$$q = -e$$

Therefore,

$$\vec F = (-e)(-2E_0\hat i)$$

$$= 2eE_0\hat i$$

Acceleration of electron :

$$a = \frac{F}{m}$$

$$= \frac{2eE_0}{m}$$

Velocity after time \(t\) :

$$v = v_0 + at$$

$$= v_0 + \frac{2eE_0t}{m}$$

de Broglie wavelength :

$$\lambda = \frac{h}{mv}$$

$$= \frac{h}{m\left(v_0+\frac{2eE_0t}{m}\right)}$$

$$= \frac{h}{mv_0+2eE_0t}$$

Given,

$$\lambda_0 = \frac{h}{4mv_0}$$

Therefore,

$$h = 4mv_0\lambda_0$$

Substituting :

$$\lambda=\frac{4mv_0\lambda_0}{mv_0+2eE_0t}$$

$$=\frac{4\lambda_0}{1+\frac{2eE_0t}{mv_0}}$$

Final Answer :

$$\lambda = \frac{4\lambda_0}{1+\frac{2eE_0t}{mv_0}}$$