$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ is:

Which of the following graphs correctly represents the de-Broglie wavelength $\lambda$ associated with a fundamental particle having linear momentum $p$?

The de-Broglie wavelength of an electron is the same as that of a $50 \ keV$ $X$-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \ MeV$ ).

The de Broglie wavelength associated with an electron accelerated through a potential difference of $\frac{200}{3} \,V$ is nearly (in $Å$)

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

What is the experimental value of the de-Broglie wavelength for an electron accelerated through a potential difference of $V$ volts?