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# The atomic number of an element whose ${k_\alpha }$ wavelength is $\lambda$ is 11. The atomic number of an element whose ${k_\alpha }$ wavelength is $4\lambda$ (A) 6(B) 11(C) 44(D) 4

Last updated date: 30th May 2024
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Answer
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Hint: Wavelength of light emitted or the wavelength of light absorbed in a transition is inversely proportional to the $z - 1$ where z is the atomic number i.e. the number of protons in the atom. In this question we are given the value of wavelength in both cases. We can find the atomic number by equating the value of constant in the 2nd equation from the 1st.

Complete step-by-step solution
According to the formula for ${k_\alpha }$
$\sqrt v = c(z - 1)$
Where v is the frequency of radiation, squaring both sides
$\dfrac{1}{\lambda } = k{(z - 1)^2}$
Where k is the constant of proportionality
Z is the atomic number of the atom
$\lambda$ is the wavelength of the atom
For the atom with atomic number 11
$\dfrac{1}{\lambda } = k{(11 - 1)^2} \\ \dfrac{1}{\lambda } = k{(10)^2} \\ \dfrac{1}{\lambda } = 100k \\$ (1)
For the atom with unknown atomic number,
$\dfrac{1}{{4\lambda }} = k{(Z - 1)^2}$ (2)
Dividing (2) by (1), we get,

$\dfrac{\lambda }{{4\lambda }} = \dfrac{{k{{(Z - 1)}^2}}}{{k(100)}}$
$\dfrac{1}{4} = \dfrac{{{{(Z - 1)}^2}}}{{(100)}}$
$\dfrac{{100}}{4} = {(Z - 1)^2}$
$5 = Z - 1 \\ Z = 6 \\$

Therefore, the correct answer is option A
Note This equation is known as Mosley’s law. Moseley was able to show that the frequencies of certain characteristic X-rays emitted from chemical elements are proportional to the square of a number which was close to the element's atomic number