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 de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy of the particle.For the regio

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy of the particle.For the region $0 \leqslant x \leqslant 1$,the potential energy is $U_1 = E_0$. Given the total energy $E = 2E_0$,the kinetic energy is $K_1 = E - U_1 = 2E_0 - E_0 = E_0$.Thus,$\lambda_1 = \frac{h}{\sqrt{2mE_0}}$.For the region $x > 1$,the potential energy is $U_2 = 0$. The kinetic energy is $K_2 = E - U_2 = 2E_0 - 0 = 2E_0$.Thus,$\lambda_2 = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4mE_0}}$.Taking the ratio $\frac{\lambda_1}{\lambda_2}$:$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE_0}}}{\frac{h}{\sqrt{4mE_0}}} = \sqrt{\frac{4mE_0}{2mE_0}} = \sqrt{2}$.

Similar Questions

If $x^2+px+1$ is a factor of the expression $ax^3+bx+c$,then

The equation of the hyperbola whose directrix is $2x + y = 1$,focus is $(1, 1)$,and eccentricity $e = \sqrt{3}$ is:

The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$ is equal to:

The products obtained in the given reactions are shown below. The number of possible products for $x$ and $y$ is:

The condition for the function $f(x)=x^3+p x^2+q x+r$ $(x \in R)$ to have no extreme value is

Vedclass Test Series

Exam Paper Generator

Online Exam Module