An electron with mass 'm' with an initial velocity $$(t = 0)$$ $$\vec{v} = v_0\hat{i}$$ $$(v_0 \gt 0)$$ enters a magnetic field $$\vec{B} = B_0\hat{j}$$. If the initial de-Broglie wavelength at $$t = 0$$ is $$\lambda_0$$ then its value after time 't' would be :

The de-Broglie wavelength of a particle is defined as
$$\lambda = \frac{h}{p} = \frac{h}{m\,v}$$
where $$v$$ is the magnitude (speed) of the velocity vector.

At $$t = 0$$ the electron has velocity $$\vec v_0 = v_0\hat i$$ and de-Broglie wavelength $$\lambda_0 = \dfrac{h}{m\,v_0}$$.

After entering the magnetic field $$\vec B = B_0\hat j$$ the electron experiences the Lorentz force
$$\vec F = q\,\vec v \times \vec B$$
For an electron, $$q = -e$$ (magnitude $$e$$). At any instant the velocity has an $$\hat i$$ component only, so
$$\vec F = (-e)\,(v_x\hat i)\times (B_0\hat j) = -e\,v_x B_0\,(\hat i \times \hat j) = -e\,v_x B_0\,\hat k$$
The force is always perpendicular to the instantaneous velocity.

Since $$\vec F \perp \vec v$$, the magnetic force does no work:
$$\text{Power} = \vec F\cdot\vec v = 0$$

Hence the kinetic energy $$\frac12 m v^2$$ remains constant, so the speed $$v$$ of the electron stays equal to its initial value $$v_0$$ for all time.

Because the speed (and therefore the magnitude of momentum $$p = m v$$) is constant, the de-Broglie wavelength also stays constant:
$$\lambda = \frac{h}{m v} = \frac{h}{m v_0} = \lambda_0$$

Thus, after any time $$t$$, the wavelength is still $$\lambda_0$$.

Final answer: $$\lambda_0$$   →   Option D