If the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is

The de Broglie wavelength of an electron moving with kinetic energy of $144 \;eV$ is nearly

$A$ bomb projected from the ground at an angle $\theta$ $\left( \theta \neq 90^\circ \right)$ explodes into two fragments of equal mass at the topmost point of its trajectory. If one of the fragments returns to the point of projection,then the ratio of the de Broglie wavelength of the second fragment just after the explosion to that of the bomb just before the explosion is:

The de Broglie wavelengths of an electron $(e)$,proton $(p)$,neutron $(n)$,and $\alpha$-particle are $\lambda_e, \lambda_p, \lambda_n$,and $\lambda_\alpha$ respectively. All have the same kinetic energy of $1 \ MeV$. Which of the following represents the correct increasing order of their wavelengths?

Consider an electron $(m = 9.1 \times 10^{-31} \ kg)$ confined by electrical forces to move between two rigid walls separated by $1.0 \times 10^{-9} \ m$,which is about five atomic diameters. The quantized energy value for the lowest stationary state is

The de Broglie wavelength associated with a proton changes by $0.25\%$. If its momentum changes by $p_0$,then its initial momentum is ......