Question
Question: When an electron in an excited state of Mo atoms falls from L to K-shell, an X-ray is emitted. These...
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∘ by planes with a separation of 2.94A∘ .What is the difference in energy between k-shell and L-shell in Mo, assuming of first order diffraction? (sin7.75∘=0.1349)
Solution
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. nλ = 2dSinθ and energy =λhc.
Complete step by step answer:
Braggs eqn. nλ = 2dSinθ and energy =λhc.
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 θ=7.75∘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.
θ=7.75∘, d= 2.94A∘ and n=1
Now, applying the Bragg's equation we get
nλ = 2dSinθ
So putting the values in this equation we will get
λ=2×2.94×Sin7.75∘
We are given that Sin7.75o = 0.1349, So replacing it we get
λ=2×2.94×0.1349, Which will give,
λ=0.712 A∘
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 λhc, putting values we get
0.712×10−10m6.626×10−34Js×3×108m/s, solving this we will get 2.79×10−15J.
Hence the energy difference between the L and K shell of Mo is 2.79×10−15J.
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.