If the kinetic energy of a free electron is doubled,its de Broglie wavelength becomes ........... times.

The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0, & 0 \le x \le 1 \\ 0, & x > 1 \end{cases}$. Let $\lambda_1$ and $\lambda_2$ be the de-Broglie wavelengths of the particle when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of the particle is $2E_0$,find $(\lambda_1/\lambda_2)^2$.

The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume $m_p = 1849 \times m_e$)

The de-Broglie wavelength of a moving bus with speed $v$ is $\lambda$. Some passengers left the bus at a stoppage. Now, when the bus moves with twice its initial speed, its kinetic energy is found to be twice its initial value. What is the de-Broglie wavelength of the bus now?

The de-Broglie wavelength $(\lambda)$ of a particle is related to its kinetic energy $(E)$ as

The wavelength of light from the spectral emission line of sodium is $589 \; nm$. Find the kinetic energy at which
$(a)$ an electron,and
$(b)$ a neutron,would have the same de Broglie wavelength.