Question

The potential energy of particle of mass $m$ varies as U(x) = \left\\{ {\begin{array}{*{20}{c}} {{E_0}\,:\,0 \leqslant x \leqslant 1} \\\ {0\,:\,\,\,\,x > 1} \end{array}} \right.
The de Broglie wavelength of the particle in the range $0 \leqslant x \leqslant 1$ is ${\lambda _1}$ and that in the range $x > 1$ is ${\lambda _2}$. If the total energy of the particle is $2{E_0}$, find $\dfrac{{{\lambda _1}}}{{{\lambda _2}}}$.
A. $\sqrt 2$
B. $2$
C. $1$
D. $\dfrac{2}{{\sqrt 2 }}$

Answer 1

The total energy and potential energy of the particle is given, using these values calculated the kinetic energy of the particle in the two given ranges separately. Recall the formula for de Broglie wavelength and use the value of kinetic energy to find the value of wavelength in the two given ranges.

Complete step by step answer:
The formula for de Broglie wavelength is given as,
$\lambda = \dfrac{h}{{\sqrt {2mK.E} }}$,
Where $h$ is the Planck’s constant, $m$ is the mass of the particle and $K.E$ is the kinetic energy of the particle.
We know total energy is the sum of potential and kinetic energy. So, using this concept we have
$E = U + K.E$ (i)

Now, we find the kinetic energy for $0 \leqslant x \leqslant 1$ and $x > 1$separately using equation (i)
For the range $0 \leqslant x \leqslant 1$ the potential energy, $U$ is ${E_0}$
Putting the value of $E$ and $U$ in equation (i), we have

\Rightarrow {\left( {K.E} \right)_1} = {E_0} $$ For the range $$x > 1$$, the potential energy , $$U$$is $$0$$ Putting the value of $$E$$ and $$U$$ in equation (i), we have $$2{E_0} = 0 + {\left( {K.E} \right)_2} \\\ \Rightarrow {\left( {K.E} \right)_2} = 2{E_0} $$ Now, for the range $$0 \leqslant x \leqslant 1$$, the de Broglie wavelength is $${\lambda _1} = \dfrac{h}{{\sqrt {2m{{\left( {K.E} \right)}_1}} }}$$ Putting the value of $${\left( {K.E} \right)_1}$$ we have, $${\lambda _1} = \dfrac{h}{{\sqrt {2m{E_0}} }}$$ (ii) Now, for the range $$x > 1$$, the de Broglie wavelength is $${\lambda _2} = \dfrac{h}{{\sqrt {2m{{\left( {K.E} \right)}_2}} }}$$ Putting the value of $${\left( {K.E} \right)_2}$$ we have, $${\lambda _2} = \dfrac{h}{{\sqrt {2m2{E_0}} }}$$ (iii) Dividing equation (ii) by (iii), we get $$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{h}{{\sqrt {2m{E_0}} }} \times \dfrac{{\sqrt {2m2{E_0}} }}{h} \\\ \therefore \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt 2 $$ **Hence, the correct answer is option A.** **Note:** Matter has dual nature that is; it has the properties of both wave and particle. When the particle is at rest it shows particle-like properties and when it is moving it shows wave-like properties. De Broglie wavelength is used to find the wavelength of the particle when it behaves as a wave.