If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

What does the de Broglie equation for an electron represent?

The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is $........$

Assuming the nitrogen molecule is moving with $r.m.s.$ velocity at $400 \ K$, the de$-$Broglie wavelength of the nitrogen molecule is close to $...... \ \mathring{A}$. (Given: nitrogen molecule mass: $4.64 \times 10^{-26} \ kg$, Boltzmann constant: $1.38 \times 10^{-23} \ J/K$, Planck constant: $6.63 \times 10^{-34} \ J \cdot s$)

The de Broglie wavelength of a particle accelerated through a potential difference of $150 \ V$ is $10^{-10} \ m$. If it is accelerated through a potential difference of $600 \ V$,what will be its wavelength in $\mathring A$?

The de Broglie wavelengths of a proton and an $\alpha$ particle are $\lambda$ and $2\lambda$ respectively. The ratio of the velocities of proton and $\alpha$ particle will be: