A) +3.4eV
B) +1.51eV
C) -3.4 eV
D) -1.51 eV
Correct Answer: B
Solution :
The ground state of hydrogen \[(n=1)\] is represented by K, the first excited state \[(n=2)\] is represented by L, the second excited state \[(n=3)\] is represented by M. So, the energy transferred by the second electron can be given by \[E=-13.6\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[=-13.6\left( \frac{1}{{{3}^{2}}}-\frac{1}{^{{{\infty }^{2}}}} \right)=-1.51\,\text{eV}\]You need to login to perform this action.
You will be redirected in
3 sec