$A$ particle is travelling $4$ times as fast as an electron. Assuming the ratio of the de-Broglie wavelength of the particle to that of the electron is $2:1$,the mass of the particle is:

The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda$. What will be its wavelength at $927^{\circ} C$?

$A$ particle of mass $1\, mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6\, m\,s^{-1}$. The velocity of the particle is (Mass of electron $= 9.1 \times 10^{-31}\, kg$)

Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would

$A$ long pipe is filled with small granules of nickel. The interatomic spacing for nickel is $d$. $A$ beam of electrons having different energies is incident on the left flat face of the pipe. Mass of electron is $m$. The beam of electrons coming out of the right flat face has energy:

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4V$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is the mass of the electron and $m_p$ is the mass of the proton).