Modern Physics Ques 153
- $K _{\alpha}$ wavelength emitted by an atom of atomic number $Z=11$ is $\lambda$. Find the atomic number for an atom that emits $K _{\alpha}$ radiation with wavelength $4 \lambda$
(a) $Z=6$
(b) $Z=4$
(c) $Z=11$
(d) $Z=44$
$(2005,2 M)$
Show Answer
Answer:
Correct Answer: 153.(a)
Solution:
Formula:
- $\quad \frac{1}{\lambda} \propto(Z-1)^{2}$
$ \therefore \quad \frac{\lambda _1}{\lambda _2}={(\frac{Z _2-1}{Z _1-1})}^{2} \text { or } \frac{1}{4}={(\frac{Z _2-1}{11-1})}^{2} $
Solving this, we get $Z _2=6$
BETA
