The $$21\,cm$$ radiowave emitted by hydrogen in interstelar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly
A.
$${10^{ - 17}}J$$
B.
$$1\,J$$
C.
$$7 \times {10^{ - 6}}J$$
D.
$${10^{ - 24}}J$$
Answer :
$${10^{ - 24}}J$$
Solution :
The energy of emitted photon is given by
$$E = \frac{{hc}}{\lambda }$$
Given, wavelength, $$\lambda = 21\,cm = 0.21\,m$$
So, $$E = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.21}}$$
$$ = {10^{ - 24}}\,J$$
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