The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of the proton,$\lambda_2$ is the wavelength associated with the photon,and the energy of the photon is $E$,then $(\lambda_1 / \lambda_2)$ is proportional to

An electron with speed $v$ and a photon with speed $c$ have the same $de-Broglie$ wavelength. If the kinetic energy and momentum of the electron are $E_{e}$ and $p_{e}$ and that of the photon are $E_{ph}$ and $p_{ph}$ respectively,which of the following is correct?

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is equal to :

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

$A$ deuteron is accelerated through $500 \ V$. For the same de Broglie wavelength,a singly ionized helium ion is accelerated through a potential difference of $V \ V$. Find the value of $V$.

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 $(\lambda_1/\lambda_2)^2$.