Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Let\[{{\lambda }_{\alpha }},{{\lambda }_{B}}\]and\[\lambda _{\alpha }^{'}\]denote the wavelength of the X-rays of the\[{{K}_{\alpha }},{{K}_{B}}\]and\[{{L}_{\alpha }}\]lines in the characteristics, X-rays for a metal: (1) \[\lambda _{\alpha }^{'}>\lambda _{\beta }^{'}>\lambda _{\alpha }^{{}}\] (2) \[\lambda _{\alpha }^{'}>\lambda _{\alpha }^{'}>\lambda _{\beta }^{{}}\] (3) \[\frac{1}{{{\lambda }_{\beta }}}=\frac{1}{{{\lambda }_{\alpha }}}+\frac{1}{\lambda _{\alpha }^{'}}\] (4) \[\frac{1}{{{\lambda }_{\alpha }}}+\frac{1}{{{\lambda }_{\beta }}}=\frac{1}{\lambda _{\alpha }^{'}}\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 and correct
D) 1, 2 and 4 are correct
Correct Answer: B
Solution :
\[{{E}_{K}}-{{E}_{L}}=\frac{hc}{{{\lambda }_{\alpha }}}\] \[{{E}_{K}}-{{E}_{M}}=\frac{hc}{{{\lambda }_{\beta }}}\] \[{{E}_{L}}-{{E}_{M}}=\frac{hc}{\lambda _{\alpha }^{'}}=\frac{hc}{\lambda _{\beta }^{{}}}-\frac{hc}{\lambda _{\alpha }^{{}}}\] Or \[\frac{1}{{{\lambda }_{\beta }}}=\frac{1}{{{\lambda }_{\alpha }}}+\frac{1}{\lambda _{\alpha }^{'}}\] Also, \[({{E}_{K}}-{{E}_{M}})>({{E}_{K}}-{{E}_{L}})>({{E}_{L}}-{{E}_{M}})\] Or \[\frac{hc}{{{\lambda }_{\beta }}}>\frac{hc}{{{\lambda }_{\alpha }}}>\frac{hc}{\lambda _{\alpha }^{'}}\]You need to login to perform this action.
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