The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is

The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda_0$. 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 $3 \times 10^6 \, m/s$. The velocity of the particle is (mass of electron $= 9.1 \times 10^{-31} \, kg$).

Two points $A$ and $B$ have potentials of $20 \ V$ and $40 \ V$ respectively. An electron is accelerated from rest between them. Find the de Broglie wavelength associated with the electron at point $B$.

$A$ proton and an alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of the proton to that of the alpha particle will be (mass of alpha particle is four times the mass of the proton,and the charge of the alpha particle is twice the charge of the proton).

An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100 \ V$. After this,their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_\alpha}$,to the nearest integer,is