A) 4.155 eV
B) 8.310 eV
C) 2.515 eV
D) 5.550 eV
Correct Answer: A
Solution :
[a] Energy of photon corresponding to first line of Balmer series \[=\left( 13.6 \right){{\left( 2 \right)}^{2}}\left[ \frac{1}{4}-\frac{1}{9} \right]\] Energy need to eject electron from n=2 level in H atom \[=\left( 13.6 \right)\left( \frac{1}{4} \right)\] So, required kinetic energy \[=\left( 13.6 \right)\left[ \left( \frac{1}{4}-\frac{1}{9} \right)-\left( \frac{1}{4} \right) \right]eV\]\[=13.6\times \left( \frac{11}{36} \right)=4.155eV\]You need to login to perform this action.
You will be redirected in
3 sec