$\begin {array} {1 1} (a)\;15.095 \: eV & \quad (b)\;1509.5\: eV \\ (c)\;150.095\: eV & \quad (d)\;150.95\: eV \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Ans : (d)

Interatomic spacing depends upon the path difference. Interatomic spacing is

$d =\large\frac{n\lambda}{2}$ and de- Broglie wave is formed between interatomic spacing.

$d_1 = \large\frac{n \lambda}{2} \: and\: d_2 = \large\frac{(n+1) \lambda}{2}$

So,$ \large\frac{ \lambda}{2} = d_2 – d_1$

$ \Rightarrow \lambda = 2(d_2 – d_1) = 2(2.5 – 2) = 1\: A^{\circ}$

$ \lambda=\large\frac{h}{p} \: \: or \: \: p = \large\frac{h}{ \lambda}$

$E = \large\frac{1}{2} mv^2$

$= \large\frac{1}{2m} (m^2v^2 ) = \large\frac{p^2}{2m}$

$ \Rightarrow E = \large\frac{h^2}{2m \lambda^2}$

$ \Rightarrow E = \large\frac{(6.6 \times 10^{-34})^2}{ [ 2 \times 9.1 \times 10^{-31} \times (10^{-10} )^2 ]} = 150.95 \: eV$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...