If $\lambda$ and $K$ are the de Broglie wavelength and kinetic energy,respectively,of a particle with constant mass,the correct graphical representation for the particle will be:

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron will be:

The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through the same potential of $100\, V$. What should nearly be the ratio of their wavelengths? $(m_{P} = 1.00727\, u, m_{e} = 0.00055\, u)$

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:

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 ..............

$A$ material particle with a rest mass $m_0$ is moving with the velocity of light $C$. Then,the wavelength of the de$-$Broglie wave associated with it is