An $\alpha$-particle moves in a circular path of radius $1 \ cm$ in a uniform magnetic field of $0.125 \ T$. The de Broglie wavelength associated with the $\alpha$-particle is

$A$ particle of mass $2 \times 10^{-27} \,kg$ has a de-Broglie wavelength of $3.3 \times 10^{-10} \,m$. The kinetic energy of this particle is (Planck's constant $h = 6.6 \times 10^{-34} \,J \cdot s$).

$A$ 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:

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with electrons change?

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