Question

A ${{K}_{\alpha }}$X-ray emitted from the sample has an energy of 7.46keV. Of which element is the sample made of
a) Calcium (Ca, Z=20)
b) Cobalt (Co, Z=27)
c) Cadmium (Cd, Z=48)
d) Nickel (Ni, Z=28)

Answer 1

Hint: The ${{K}_{\alpha }}$X-ray emitted from the sample represents the transition of an electron of an atom from the second energy state to the first energy state. When an electron makes a transition from higher to lower energy state it emits energy in the form of radiations. The energy emitted by the electron during the transition corresponds to $h\gamma $ where h is the Planck’s constant and \[\gamma \]is the frequency of the emitted radiation. The energy of an electron in the nth energy state is given by ${{E}_{n}}=\dfrac{-13.6{{\left( Z \right)}^{2}}}{{{n}^{2}}}$. Hence we can find the energy emitted by the electron by taking the difference between their energy states and equating it to the given energy of the X-rays radiation to determine the atomic number Z of the element.

Complete answer:
When electrons get de-excited, they release whatever energy they absorb in the form of radiation. This radiation emitted has a particular frequency. The energy of the emitted radiation is equal to the difference between the energy states of the permitted orbits of the electrons. When an electrons makes a transition from n energy state to N energy state, the energy released in the form of radiation is given by,
$\begin{align}
  & {{E}_{N}}-{{E}_{n}}=\dfrac{-13.6{{\left( Z \right)}^{2}}}{{{N}^{2}}}-\dfrac{-13.6{{\left( Z \right)}^{2}}}{{{n}^{2}}} \\
 & {{E}_{N}}-{{E}_{n}}=\dfrac{13.6{{\left( Z \right)}^{2}}}{{{n}^{2}}}-\dfrac{13.6{{\left( Z \right)}^{2}}}{{{N}^{2}}} \\
\end{align}$
It is given in the question that the energy emitted is numerically equal to 7.46keV the electron makes a transition from the second energy state to the first energy state. Hence the atomic number Z from the above equation is,
$\begin{align}
  & \Rightarrow {{E}_{2}}-{{E}_{1}}=\dfrac{13.6{{\left( Z \right)}^{2}}}{{{1}^{2}}}-\dfrac{13.6{{\left( Z \right)}^{2}}}{{{2}^{2}}} \\
 & \Rightarrow {{E}_{2}}-{{E}_{1}}=\dfrac{13.6{{\left( Z \right)}^{2}}}{{{1}^{2}}}-\dfrac{13.6{{\left( Z \right)}^{2}}}{{{2}^{2}}} \\
 & \Rightarrow 7.46=13.6{{\left( Z \right)}^{2}}\dfrac{\left[ 4-1 \right]}{4} \\
 & \Rightarrow 4\times 7.46\times {{10}^{3}}=13.6\times 3{{\left( Z \right)}^{2}} \\
 & \Rightarrow {{\left( Z \right)}^{2}}=731.37 \\
 & \Rightarrow Z=27 \\
\end{align}$

So, the correct answer is “Option B”.

Note:
When an electron in the lower energy state gains energy, it gets excited and jumps to the higher energy state. But electrons in the higher energy state are always unstable. Hence they jump back to their lower energy state and emit radiation corresponding to a particular frequency.