Dual Nature Of Light Question 115
Question: Let $ {\lambda_{\alpha }} $ , $ {\lambda_{\beta }} $ and $ {{{\lambda }’}{\alpha }} $ denote the wavelengths of the X-rays of the $ {K{\alpha }},,{K_{\beta }} $ and $ {L_{\alpha }} $ lines in the characteristic X-rays for a metal
Options:
A) $ {\lambda_{\alpha }}>{{{\lambda }’}{\alpha }}>{\lambda{\beta }} $
B) $ {{{\lambda }’}{\alpha }}>{\lambda{\beta }}>{\lambda_{\alpha }} $
C) $ \frac{1}{{\lambda_{\beta }}}=\frac{1}{{\lambda_{\alpha }}}+\frac{1}{{{{{\lambda }’}}_{\alpha }}} $
D) $ \frac{1}{{\lambda_{\alpha }}}+\frac{1}{{\lambda_{\beta }}}=\frac{1}{{{{{\lambda }’}}_{\alpha }}} $
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Answer:
Correct Answer: C
Solution:
According to the energy diagram of X-ray spectra $ \because \Delta E=\frac{hc}{\lambda } $
$ \Rightarrow \lambda \propto \frac{1}{\Delta E} $ (DE = Energy radiated when e? jumps from, higher energy orbit to lower energy orbit) $ \because {{(\Delta E)}{{k{\beta }}}}>{{(\Delta E)}{{k{\alpha }}}}>{{(\Delta E)}{{L{\alpha }}}} $
$ \therefore {{{\lambda }’}{\alpha }}>{\lambda{\alpha }}>{\lambda_{\beta }} $ Also $ {{(\Delta E)}{{k{\beta }}}}={{(\Delta E)}{{k{\alpha }}}}+{{(\Delta E)}{{L{\alpha }}}} $
$ \Rightarrow \frac{hc}{{\lambda_{\beta }}}=\frac{hc}{{\lambda_{\alpha }}}+\frac{hc}{{{{{\lambda }’}}{\alpha }}} $
$ \Rightarrow \frac{1}{{\lambda{\beta }}}=\frac{1}{{\lambda_{\alpha }}}+\frac{1}{{{{{\lambda }’}}_{\alpha }}} $
BETA
