A) 3
B) 4
C) 5
D) 2
Correct Answer: A
Solution :
\[E=K{{E}_{\max }}+W\] \[=e{{V}_{0}}+W\] \[=10+2.75\] \[E=12.75\,eV\] Difference of 4 and 1 energy level is 12.75 eV. So, higher energy level is 4 to ground and excited state is n = 3.You need to login to perform this action.
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