Question: 26. The graph which shows the variation of \left(\frac{1}{\lambda^{2}}\right) and its kinetic energy, E is (where \lambda is de Broglie wavelength of a free particle) :
Answer: Option (4)
Explanation:
The de Broglie wavelength of a free particle is given by
\lambda = \frac{h}{p}For a non-relativistic particle, momentum is related to kinetic energy by
p = \sqrt{2mE}Substituting this expression for momentum,
\lambda = \frac{h}{\sqrt{2mE}}Taking reciprocal and squaring both sides,
\frac{1}{\lambda^{2}} = \frac{2mE}{h^{2}}This equation shows that \frac{1}{\lambda^{2}}
is directly proportional to the kinetic energy E.
Therefore, the graph between \frac{1}{\lambda^{2}}
and E is a straight line passing through the origin.
Hence, the correct graph is represented by Option (4).