The potential energy of a particle of mass m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The total energy of the particle is $E = 2 E_0$.For the region $0 \le x \le 1$,the potential energy is $U_1 = E_0$. The kinetic energy is $K_1 = E

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) The total energy of the particle is $E = 2 E_0$.For the region $0 \le x \le 1$,the potential energy is $U_1 = E_0$. The kinetic energy is $K_1 = E - U_1 = 2 E_0 - E_0 = E_0$.The de-Broglie wavelength is $\lambda_1 = \frac{h}{\sqrt{2m K_1}} = \frac{h}{\sqrt{2m E_0}}$.For the region $x > 1$,the potential energy is $U_2 = 0$. The kinetic energy is $K_2 = E - U_2 = 2 E_0 - 0 = 2 E_0$.The de-Broglie wavelength is $\lambda_2 = \frac{h}{\sqrt{2m K_2}} = \frac{h}{\sqrt{2m(2 E_0)}} = \frac{h}{\sqrt{4m E_0}}$.Taking the ratio: $\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2m E_0}}{h / \sqrt{4m E_0}} = \sqrt{\frac{4m E_0}{2m E_0}} = \sqrt{2}$.

Similar Questions

For which of the following particles will it be most difficult to experimentally verify the de Broglie relationship?

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times 10^{-12} \ m$, the minimum electron energy required is close to .............. $keV$.

The de Broglie wavelength of an electron moving with a velocity $1.5 \times 10^8 \, m/s$ is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is

$A$ proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is $2000$ times heavier than an electron,what will be the relation between the de Broglie wavelength of the proton $(\lambda_{p})$ and that of the electron $(\lambda_{e})$?

What is the theoretical formula for the de-Broglie wavelength of a particle?

Vedclass Test Series

Exam Paper Generator

Online Exam Module