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

The de-Broglie wavelength of a particle of kinetic energy $K$ is $\lambda$. What will be the wavelength of the particle if its kinetic energy becomes $\frac{K}{4}$?

An electron of mass $m$ is accelerated through a potential difference of $V$. Its de Broglie wavelength is $\lambda$. If a proton of mass $M$ is accelerated through the same potential difference,its de Broglie wavelength will be ..............

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field,the radius of the $n^{th}$ orbital will therefore be proportional to

The magnitude of the de-Broglie wavelength $(\lambda)$ of electron $(e)$,proton $(p)$,neutron $(n)$ and $\alpha-$ particle $(\alpha)$ all having the same energy of $1\,MeV$,in the increasing order will follow the sequence

$A$ particle $P$ is formed due to a completely inelastic collision of particles $x$ and $y$ having de-Broglie wavelengths $\lambda_x$ and $\lambda_y$ respectively. If $x$ and $y$ were moving in opposite directions,then the de-Broglie wavelength of $P$ is