The de Broglie wavelength of an electron of kinetic energy $9 \ eV$ is (take $h=4 \times 10^{-15} \ eV \cdot s$,$c=3 \times 10^{10} \ cm/s$ and the mass $m_e$ of electron as $m_e c^2=0.5 \ MeV$)

The velocity of a particle $A$ is $3$ times the velocity of a proton. If the ratio of the de Broglie wavelengths of the particle $A$ and the proton is $3:2$,the mass of the particle $A$ is (where $m_{p}$ is the mass of the proton).

$A$ particle of mass $1 \times 10^{-30} \,kg$ and electric charge $1.6 \times 10^{-19} \,C$ has a de-Broglie wavelength of $660 \,nm$. The kinetic energy of this particle is (Planck's constant,$h = 6.6 \times 10^{-34} \,J \cdot s$)

If the momentum of an electron changes by $P$,then the de-Broglie wavelength associated with it changes by $5 \%$. The initial momentum of the electron is: (in $P$)

$A$ 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).

Two electrons are moving with the same speed $v$. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field. After some time,if the de-Broglie wavelengths of the two are $\lambda_1$ and $\lambda_2$,then: