Question

When an electron in an excited state of Mo atoms falls from L to K-shell, an X-ray is emitted. These X-rays are diffracted at angle of \[{7.75^ \circ }\] by planes with a separation of \[2.94{A^ \circ }\] .What is the difference in energy between k-shell and L-shell in Mo, assuming of first order diffraction? \[(sin{7.75^ \circ } = 0.1349)\]

Answer 1

Hint: We can use Bragg's equation to know the wavelength of X-Ray produce and use the energy equation to know the energy difference.
Formula Used:
Braggs eqn. ${\text{n}}\lambda {\text{ = 2dSin}}\theta $ and energy =$\dfrac{{{\text{hc}}}}{\lambda }$.

Complete step by step answer:
Braggs eqn. ${\text{n}}\lambda {\text{ = 2dSin}}\theta $ and energy =$\dfrac{{{\text{hc}}}}{\lambda }$.
Complete Step by step answer:
To solve this question first we need to know what all is given to us, we are provided with the atom i.e. Mo along with a diffracted angle of X-Ray emitted which is \[\theta = {7.75^ \circ }\]by planes which have a separation of 2.94 A. Also, we have to assume that it’s a first order diffraction. So all this information is provided to us.
\[\theta = {7.75^ \circ }\], d= \[2.94{A^ \circ }\] and n=1
Now, applying the Bragg's equation we get
${\text{n}}\lambda {\text{ = 2dSin}}\theta $
So putting the values in this equation we will get
$\lambda = 2 \times 2.94 \times {\text{Sin7}}{\text{.7}}{{\text{5}}^ \circ }$
We are given that $Sin7.75^o$ = 0.1349, So replacing it we get
$\lambda = 2 \times 2.94 \times 0.1349$, Which will give,
$\lambda = 0.712{\text{ }}{{\text{A}}^ \circ }$
Now we are having the wavelength of X-Ray which is emitted when the atom falls from the L to K shell. To find the energy difference we will use the formula $\dfrac{{{\text{hc}}}}{\lambda }$, putting values we get
$\dfrac{{6.626 \times {{10}^{ - 34}}{\text{Js}} \times 3 \times {{10}^8}{\text{m/s}}}}{{0.712 \times {{10}^{ - 10}}{\text{m}}}}$, solving this we will get \[2.79 \times {10^{ - 15}}J\].
Hence the energy difference between the L and K shell of Mo is \[2.79 \times {10^{ - 15}}J\].
Note:
We must know that the K shell is the shell which contains only s subshell and it is the first shell present in any atom, also this shell is the nearest to the nucleus of the atom and contains only 2 electrons at max. The L shell contains two subshells which are s and p respectively; also it is the second shell after the K shell and contains a maximum of 8 electrons, two in s subshell and six in p subshell.