If the momentum of an electron is changed by $\Delta P,$ then the de Broglie wavelength associated with it changes by $0.5\%.$ The initial momentum of the electron will be

$A$ particle of mass $1\, mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6\, m\,s^{-1}$. The velocity of the particle is (Mass of electron $= 9.1 \times 10^{-31}\, kg$)

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 wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon of same energy $E$ are related by

$A$ proton and an alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of the proton to that of the alpha particle will be (mass of alpha particle is four times the mass of the proton,and the charge of the alpha particle is twice the charge of the proton).

An object of mass $200 \ g$ is moving with a speed of $5 \ m/hr$. The order of the de Broglie wavelength associated with it is ...... $(h = 6.626 \times 10^{-34} \ J \cdot s)$.