Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?
A.
B.
C.
D.
Answer :
Solution :
The de-Broglie wavelength is given by
$$\lambda = \frac{h}{P} \Rightarrow P\lambda = h$$
This equation is in the form of $$yx = c,$$ which is the equation of a rectangular hyperbola. Hence, the graph given in option (B) is the correct one.
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