The wavelength of the matter wave is independent of

An electron beam,when accelerated by a voltage of $10 \ kV$,has a de-Broglie wavelength of $\lambda$. If the voltage is increased to $20 \ kV$,then the de-Broglie wavelength associated with the electron beam would be:

The ratio of wavelengths of a proton and a deuteron accelerated by potentials $V_{p}$ and $V_{d}$ is $1 : \sqrt{2}$. Then,the ratio of $V_{p}$ to $V_{d}$ will be:

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 \,nm$ to $0.5 \,nm$ is

Two particles move at right angles to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles undergo a perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle is given by

The ratio of the de-Broglie wavelengths for the electron and proton moving with the same velocity is ($m_p$ = mass of proton,$m_e$ = mass of electron).