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

If ${\lambda _p}$ and ${\lambda _\alpha }$ are the de-Broglie wavelengths associated with protons and $\alpha$-particles of equal kinetic energies,then:

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: $c = 3 \times 10^8 \ m/s$)

The de Broglie wavelength associated with a proton changes by $0.25\%$. If its momentum changes by $p_0$,then its initial momentum is ......

The ratio of de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is