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

Of the following particles having the same kinetic energy,the one which has the largest de Broglie wavelength is

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is

The ratio of the de-Broglie wavelengths for the electron and proton moving with the same velocity is ($m_p$ = mass of proton,$m_e$ = mass of electron).

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ 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?