Question 54: Dual Nature of Matter and Radiation

Question

The potential energy of a particle of mass m 
is given by
$$\mathrm{U}(x)=\left\{\begin{array}{cc}\mathrm{E}_{0} & 0 \leq x \leq 1 \ 0 & x>1\end{array}ight.$$
$$\lambda_{1}$$ and $$\lambda_{2}$$ are the de Broglie wavelengths of the particle, when $$0 \leq x \leq 1$$ and $$x > 1$$, respectively. If the total energy of particle is $$2 \mathrm{E}_{0}$$, find $$\frac{\lambda_{1}}{\lambda_{2}}$$.

Accepted Answer

$$U(x) = \left\{ {\matrix{
   {{E_0}} & {0 \le x \le 1}  \cr 
   0 & {x > 1}  \cr

} } \right.$$
$$K.E = \left\{ {\matrix{
   {2{E_0} - {E_0} = {E_0}} & {0 \le x \le 1}  \cr 
   {2{E_0} - 0 = 2{E_0}} & {x > 1}  \cr

} } \right.$$
From $$\lambda  = {h \over {\sqrt {2mE} }}$$
$${\lambda _1} = {h \over {\sqrt {2m{E_0}} }}$$ and $${\lambda _2} = {h \over {\sqrt {2m{E_0}} }} = {h \over {\sqrt {4m{E_0}} }}$$
Now $${{{\lambda _1}} \over {{\lambda _2}}} = {{{h \over {\sqrt {2m{E_0}} }}} \over {{h \over {\sqrt {4m{E_0}} }}}} = {{\sqrt {4m{E_0}} } \over {\sqrt {2m{E_0}} }} = \sqrt {{{4m{E_0}} \over {2m{E_0}}}} \sqrt 2 $$
$${{{\lambda _1}} \over {{\lambda _2}}} = \sqrt 2 $$

Answer

2

Answer

$$\sqrt2$$

Answer

$$\sqrt3$$

Answer

3

Dual Nature of Matter and Radiation

Choose the correct answer: