A) 3
B) 2
C) 6
D) 10
Correct Answer: B
Solution :
Energy provided to the ground state electron |
\[=\frac{hc}{\lambda }=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{975\times {{10}^{-10}}}\] |
\[=\frac{6.6\times 3}{975}\times {{10}^{-16}}\] |
\[=0.020\times {{10}^{-16}}=2\times {{10}^{-18}}J\] |
\[=\frac{20\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\] |
\[=\frac{20}{1.6}eV=12.5eV\] |
It means the electron is jump to \[n=3\] from \[n=1\]. The number of field lines possible from \[n=3\] |
To \[n=1\] is 2. |
\[n\to 3,\] to \[n\to 2\] to \[n\to 1\] |
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