An $\alpha$-particle moves in a circular path of radius $0.83\, cm$ in the presence of a magnetic field of $0.25\, Wb/m^2$. The de Broglie wavelength associated with the particle will be .............. $\mathring{A}$.

The de-Broglie wavelength associated with a particle of mass $m$ moving with velocity $v$ is:

Electrons are accelerated through a potential difference of $16 \ kV$. If the potential difference is increased to $64 \ kV$,then the de-Broglie wavelength associated with the electron will

According to the de-Broglie hypothesis,the wavelength associated with a moving electron of mass $m$ is $\lambda_e$. Using the mass-energy relation and Planck's quantum theory,the wavelength associated with a photon is $\lambda_p$. If the energy $(E)$ of the electron and the photon is the same,then the relation between $\lambda_e$ and $\lambda_p$ is:

An electron of mass $m$ has de-Broglie wavelength $\lambda$ when accelerated through potential difference $V$. When a proton of mass $M$ is accelerated through a potential difference of $9V$,the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage).

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is: