Question

Assume that the de-Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance ${}^{\prime }{d}^{\prime }$ between the atoms of the array is $2A^{\circ }$. A similar standing wave is again formed if ${}^{\prime }{d}^{\prime }$ is increased to $2.5A^{\circ }$ but not for any intermediate value of ${}^{\prime }{d}^{\prime }$. Find the energy of the electron in electron volts and the least value of ${}^{\prime }{d}^{\prime }$ for which the standing wave of the type described above can form.

• A. $151eV,0.5A^{\circ }$.
• B. $1151eV,0.5A^{\circ }$.
• C. $251eV,0.5A^{\circ }$.
• D. $1351eV,0.5A^{\circ }$.

Answer 1

The correct option is A $151eV,0.5A^{\circ }$.

As nodes are formed at each of the atomic sites, hence

$2\stackrel{0}{A}=n\left(\lambda /2\right)⟶\left(1\right)$

and $2.5\stackrel{0}{A}=(n+1)\lambda /2$

$\therefore$ $2.5/2=nt1/n.5/4=n+1/n$ or $n=4$

Hence from equation $\left(1\right)$

$2\stackrel{0}{A}=4\lambda /2$ i.e. $\lambda =1\stackrel{0}{A}$

Now, de Brogile wavelength is given by,

$\lambda =h/\sqrt{2}mk$ or $k=h^2/{\lambda }^2=2m$

$\therefore$ $k={(6.63\times {10}^{-34})}^2/{(1\times {10}^{-10})}^2\times 2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}eV$

$k=151eV$

$\theta$ will be minimum, when

$n=1$, $d_{min}=\lambda /2=1\stackrel{0}{A}/2=0.5\stackrel{0}{A}$