If the momentum of electron is changed by $$P,$$ then the de Broglie wavelength associated with it changes by $$0.5\% .$$ The initial momentum of electron will be:
A.
$$200\,P$$
B.
$$400\,P$$
C.
$$\frac{P}{{200}}$$
D.
$$100\,P$$
Answer :
$$200\,P$$
Solution :
The de-Broglie’s wavelength associated with the moving electron $$\lambda = \frac{h}{P}$$
Now, according to problem
$$\eqalign{
& \frac{{d\lambda }}{\lambda } = - \frac{{dp}}{P}; \cr
& \frac{{0.5}}{{100}} = \frac{P}{{P'}} \cr
& \Rightarrow P' = 200\,P \cr} $$
Releted MCQ Question on Modern Physics >> Dual Nature of Matter and Radiation
Releted Question 1
A particle of mass $$M$$ at rest decays into two particles of
masses $${m_1}$$ and $${m_2},$$ having non-zero velocities. The ratio of the de Broglie wavelengths of the particles, $$\frac{{{\lambda _1}}}{{{\lambda _2}}},$$ is
A proton has kinetic energy $$E = 100\,keV$$ which is equal to that of a photon. The wavelength of photon is $${\lambda _2}$$ and that of proton is $${\lambda _1}.$$ The ration of $$\frac{{{\lambda _2}}}{{{\lambda _1}}}$$ is proportional to