The de-Broglie wavelength $\lambda$ associated with an electron having kinetic energy $E$ is given by the expression:

If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

Two particles have the same charge. If they are accelerated through the same potential difference,what will be the ratio of their de Broglie wavelengths?

The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0, & 0 \le x \le 1 \\ 0, & x > 1 \end{cases}$. Let $\lambda_1$ and $\lambda_2$ be the de-Broglie wavelengths of the particle when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of the particle is $2E_0$,find $(\lambda_1/\lambda_2)^2$.

If the potential difference used to accelerate electrons is increased four times,by what factor does the de-Broglie wavelength associated with the electrons change?

If we consider an electron and a photon of the same de-Broglie wavelength,then they will have the same: