Question

The wavelengths of \[{K_\alpha }\] X-rays for lead isotopes \[P{b^{208}}\], \[P{b^{206}}\] and \[P{b^{204}}\] are \[{\lambda _1}\], \[{\lambda _2}\] and \[{\lambda _3}\], then:
(A) \[{\lambda _2} = \sqrt {{\lambda _1}{\lambda _3}} \]
(B) \[{\lambda _2} = {\lambda _1} + {\lambda _3}\]
(C) \[{\lambda _2} = {\lambda _1}{\lambda _3}\]
(D) \[{\lambda _2} = \dfrac{{{\lambda _1}}}{{{\lambda _3}}}\]

Answer 1

Hint: It is given that wavelength of \[{K_\alpha }\] line for isotopes of lead said to be \[{\lambda _1}\],\[{\lambda _2}\] and \[{\lambda _3}\] respectively . Using Mosley’s law, find the ratio between the wavelength of the particle using Mosley’s law equation and find the unknown value.

Complete step by step solution:
 Moseley’s law states that the frequency of any spectral line in the X-ray spectrum is directly proportional to the square of the atomic number of the emitting element. Using this law we can find the atomic number of the unknown element. Moseley’s law has helped in discovering elements such as Technetium, Rhodium etc.
Mathematically, we can represent this formula as,
\[ \Rightarrow \upsilon = {(a(z - b))^2}\], Where a and b are constants and z is atomic number of the element.
 Now, we know that frequency is given as the ratio of speed of light to the wavelength of the particle.
\[ \Rightarrow \dfrac{c}{\lambda } = a{(z - b)^2}\]
From this equation, we can conclude that wavelength of the particle is inversely proportional to the square of the atomic number of the particle. This is represented as,
\[\dfrac{1}{\lambda } \propto a{(z - b)^2}\]
Now, for our first case, we know that the atomic number of the particle is 43. Thus we take out the proportionality symbol, by using a constant term R. R is known as Rydberg’s constant. Now, the equation changes to
\[ \Rightarrow \dfrac{1}{{{\lambda _1}}} = R{(a(82 - b))^2}\]
For the first isotope, the value of b is equal to 1.

\[ \Rightarrow \dfrac{1}{{{\lambda _1}}} = R{(a(82 - 1))^2}\]-----(1)

Now applying the same for second and third isotope we get,
\[ \Rightarrow \dfrac{1}{{{\lambda _2}}} = \dfrac{1}{{{\lambda _3}}} = R{(a(82 - 1))^2}\]---(2) (Atomic number never changes for isotopes)
 Now, on carefully observing the equations we can say that,
\[ \Rightarrow {(\dfrac{1}{{{\lambda _2}}})^2} = \dfrac{1}{{{\lambda _3}}} \times \dfrac{1}{{{\lambda _1}}}\]
Multiplying and reciprocating, we get
\[ \Rightarrow {\lambda _2}^2 = {\lambda _3}{\lambda _1}\]
Taking Square root value we get,
\[ \Rightarrow {\lambda _2} = \sqrt {{\lambda _3}{\lambda _1}} \]

Thus, option (A) is the right answer for the given question.

Note: In the emission spectrum, K-alpha lines are wavelength lines that correspond to radiation emission when an electron transitions from the nearest shell of specified quantum number to the innermost shell or kth shell.