A nucleus of mass M at rest splits into two p | vedclass

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

Question

(C) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the total initial momentum is zero.Therefore,the magn

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

(C) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the total initial momentum is zero.Therefore,the magnitudes of the momenta of the two parts must be equal and opposite:$|\vec{P}_1| = |\vec{P}_2| = P$The de Broglie wavelength $\lambda$ is given by the relation:$\lambda = \frac{h}{P}$Since both parts have the same magnitude of momentum $P$,their de Broglie wavelengths will also be equal.Therefore,the ratio of their de Broglie wavelengths is $\lambda_1 : \lambda_2 = 1 : 1$.

$A$ nucleus of mass $M$ at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2M'}{3}$ (where $M' < M$). The ratio of the de Broglie wavelengths of the two parts will be:

Similar Questions

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

Write a note on the electron microscope.

If a ball of mass $1 \ kg$ has a velocity of $1 \ m/s$,what is its de Broglie wavelength?

The de-Broglie wavelength of a particle of kinetic energy $K$ is $\lambda$. What will be the wavelength of the particle if its kinetic energy becomes $\frac{K}{4}$?

$A$ proton and an electron are accelerated through the same potential difference. The ratio $\frac{\lambda_e}{\lambda_p}$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module