Search any question & find its solution
Question:
Answered & Verified by Expert
An element with atomic number $Z=11$ emits $K_{\alpha}-$ X-ray of wavelength $\lambda$. The atomic number which emits $K_{\alpha}$-X-ray of wavelength $4 \lambda$ is
Options:
Solution:
1470 Upvotes
Verified Answer
The correct answer is:
6
According to Moseley's law $\sqrt{\mathrm{v}}=\mathrm{a}(\mathrm{z}-\mathrm{b})$ or $\mathrm{v}=\mathrm{a}^{2}(\mathrm{z}-\mathrm{b})^{2}$
or $\frac{\mathrm{c}}{\lambda}=\mathrm{a}^{2}(\mathrm{z}-\mathrm{b})^{2}$
$\therefore \frac{\lambda_{1}}{\lambda_{2}}=\frac{\left(\mathrm{z}_{2}-1\right)^{2}}{\left(\mathrm{z}_{1}-1\right)^{2}}$
Here $\lambda_{1}=\lambda, \lambda_{2}=4 \lambda, \mathrm{z}_{1}=11$ and $\mathrm{z}_{2}=$ ?
$\therefore \frac{1}{4 \lambda}=\frac{\left(z_{2}-1\right)^{2}}{(11-1)^{2}}$
or $\left(\mathrm{z}_{2}-1\right)^{2}=25$ or $\mathrm{z}_{2}=6$
or $\frac{\mathrm{c}}{\lambda}=\mathrm{a}^{2}(\mathrm{z}-\mathrm{b})^{2}$
$\therefore \frac{\lambda_{1}}{\lambda_{2}}=\frac{\left(\mathrm{z}_{2}-1\right)^{2}}{\left(\mathrm{z}_{1}-1\right)^{2}}$
Here $\lambda_{1}=\lambda, \lambda_{2}=4 \lambda, \mathrm{z}_{1}=11$ and $\mathrm{z}_{2}=$ ?
$\therefore \frac{1}{4 \lambda}=\frac{\left(z_{2}-1\right)^{2}}{(11-1)^{2}}$
or $\left(\mathrm{z}_{2}-1\right)^{2}=25$ or $\mathrm{z}_{2}=6$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.