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12.Atoms
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ક્ષ કિરણ ટ્યૂબ વડે ઉત્પન્ન થતાં $K_\alpha$ યક્ષ કિરણની તરંગ લંબાઈ $ 0.76\, \mathring A $ છે. ટ્યૂબના એનોડ પદાર્થનો પરમાણ્વિય આંક ......છે.
A$20$
B$60$
C$41$
D$80$
Solution
જેમ આપણે જાણીએ છીએ,
$\frac{1}{{{\lambda _{K\alpha }}}} = R{(Z – 1)^2}\left[ {\frac{1}{{n_1^2}} – \frac{1}{{n_2^2}}} \right]\,\,$
જ્યાં ${{\text{n}}_{\text{1}}} = 1,\,\,{n_2} = 2$
$\frac{1}{{{\lambda _{K\alpha }}}} = R{(Z – 1)^2}\left( {\frac{3}{4}} \right),$
$ \Rightarrow \,\,{(Z – 1)^2} = \frac{4}{{3{\lambda _{K\alpha }}R}}$
${(Z – 1)^2} = \frac{4}{{3 \times 0.76 \times {{10}^{ – 10}} \times 1.097 \times {{10}^7}}}\,$
$\,\, \Rightarrow \,\,\,\,Z – 1 = 40\,\,\,\,$
$\therefore$$\,\,\,Z = 41$
$\frac{1}{{{\lambda _{K\alpha }}}} = R{(Z – 1)^2}\left[ {\frac{1}{{n_1^2}} – \frac{1}{{n_2^2}}} \right]\,\,$
જ્યાં ${{\text{n}}_{\text{1}}} = 1,\,\,{n_2} = 2$
$\frac{1}{{{\lambda _{K\alpha }}}} = R{(Z – 1)^2}\left( {\frac{3}{4}} \right),$
$ \Rightarrow \,\,{(Z – 1)^2} = \frac{4}{{3{\lambda _{K\alpha }}R}}$
${(Z – 1)^2} = \frac{4}{{3 \times 0.76 \times {{10}^{ – 10}} \times 1.097 \times {{10}^7}}}\,$
$\,\, \Rightarrow \,\,\,\,Z – 1 = 40\,\,\,\,$
$\therefore$$\,\,\,Z = 41$
Standard 12
Physics
