A) \[{{10}^{3}}\,\,\pi J\]
B) \[V={{I}_{b}}{{R}_{b}}\]
C) \[\Rightarrow \]
D) \[{{R}_{b}}=\frac{V}{{{I}_{b}}}\]
Correct Answer: D
Solution :
The energy expression in terms, of Rydberg?s, Planck?s constant \[h\] and velocity of light \[c\] is \[E=-{{Z}^{2}}\frac{Rhc}{{{n}^{2}}}\] Where \[Z\] is atomic number. The expression on keeping the values of constants, reduces to \[E=-{{Z}^{2}}\frac{13.6}{{{n}^{2}}}eV\] Where \[Z=1\] \[\left( hydrogen \right)\], \[{{n}_{1}}=1,\,\,{{n}_{2}}=2\], we have \[{{E}_{1}}=-{{\left( 1 \right)}^{2}}\times 13.6\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] \[=-13.6\left( 1-\frac{1}{4} \right)\] ?.(1) When \[Z=10,\,\,{{n}_{1}}=1,\,{{n}_{2}}=2\], we have \[{{E}_{2}}=-{{\left( 10 \right)}^{2}}\times 13.6\left( \frac{1}{1}-\frac{1}{{{2}^{2}}} \right)\] ?(2) Dividing \[Eq.\,\left( 1 \right)\,\,by\,\,Eq.\,\,\left( 2 \right),\,we\,get\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{1}{100}\] \[\Rightarrow \] \[{{E}_{2}}=100\times {{E}_{1}}\] Given, \[{{E}_{1}}=10.2\,\,eV\] \[\therefore \] \[{{E}_{2}}=100\times 10.2=1020\,\,eV\]
You need to login to perform this action.
You will be redirected in
3 sec
