If the kinetic energy of the particle is increased by $16$ times,the percentage change in the de Broglie wavelength of the particle is ............$\%$

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

Write the equation for the de-Broglie wavelength of a particle having momentum $(p)$.

The de-Broglie wavelength of an electron having $80 \ eV$ of energy is nearly .............. $\mathring{A}$ ($1 \ eV = 1.6 \times 10^{-19} \ J$,Mass of electron $= 9 \times 10^{-31} \ kg$,Planck's constant $= 6.6 \times 10^{-34} \ J \cdot s$).

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$. If both particles move along the $x$-axis after the collision,the change $\Delta \lambda$ in the de-Broglie wavelength of particle $A$,in terms of its de-Broglie wavelength $(\lambda_{0})$ before the collision is:

$A$ particle is moving along the $x-$axis back and forth in a box of length $L$. Assuming the de-Broglie hypothesis is applicable for the particle,and it is moving with a constant speed in the box making perfectly elastic collisions with the walls. The possible value of the momentum of the particle is (Note: $h$ is Planck's constant,$n$ is the principal quantum number).