A) \[{{n}_{4}}\xrightarrow{{}}{{n}_{1}}\]
B) \[{{n}_{4}}\xrightarrow{{}}{{n}_{3}}\]
C) \[{{n}_{5}}\xrightarrow{{}}{{n}_{4}}\]
D) \[{{n}_{5}}\xrightarrow{{}}{{n}_{1}}\]
Correct Answer: B
Solution :
[b]| The energy associated with radiation of wave length \[975\text{ }\overset{o}{\mathop{\Alpha }}\,\]is given by \[E=\frac{12400}{975}eV=12.72eV\] |
| [for a radiation of wavelength x\[\overset{o}{\mathop{A}}\,\], we know\[E=\frac{12400}{x}eV\]] |
| \[E=13.6\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]eV\] |
| Or \[E=13.6\left[ \frac{1}{1}-\frac{1}{n_{2}^{2}} \right]eV\left[ \because {{n}_{1}}=1 \right]\] |
| Or \[12.72=13.6\left[ 1-\frac{1}{n_{2}^{2}} \right]\left[ \because E=12.72eV \right]\] |
| Or \[1-\frac{1}{n_{2}^{2}}=\frac{12.72}{13.6}\] |
| Or \[\frac{1}{n_{2}^{2}}=\frac{13.6-12.72}{13.6}=\frac{0.88}{13.6}=\frac{1}{16}\] |
| Or \[n_{2}^{2}=16\,\,or\,\,{{n}_{2}}=4\] |
| The transition \[{{n}_{4}}\xrightarrow{{}}{{n}_{3}}\]will give the longest wave length. |
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