A) 3
B) 2
C) 6
D) 10
Correct Answer: A
Solution :
Energy provided to the ground state electron \[=\frac{hc}{\lambda }=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{975\times {{10}^{-10}}}\] \[=\frac{6.6\times 3}{975}\times {{10}^{-16}}\] \[=0.020\times {{10}^{-16}}=2\times {{10}^{-18}}\text{J}\] \[=\frac{20\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\] \[=\frac{20}{1.6}eV=12.5eV\] 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 ® 3, to n ® 2 to n ® 1.
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