A particle of mass $$m$$ is projected from ground with velocity making angle $$\theta $$ with the vertical. The de-Broglie wavelength of the particle at the highest point is -
A.
$$\infty $$
B.
$$\frac{h}{{mu\sin \theta }}$$
C.
$$\frac{h}{{mu\cos \theta }}$$
D.
$$\frac{h}{{mu}}$$
Answer :
$$\frac{h}{{mu\sin \theta }}$$
Solution :
Velocity at highest point $$ = u\sin \theta .$$
$$\therefore {\lambda _D} = \frac{h}{{mu\sin \theta }}\,\,\left( {{\text{Since }}\theta {\text{ is velocity wrt vertical}}} \right)$$
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