Classification Of Elements And Periodicity In Properties Question 26
Question: If first orbit energy of $ H{{e}^{+}} $ is - 54.4 eV, then the second orbit energy will be [Roorkee 1995]
Options:
A) - 54.4 eV
B) - 13.6 eV
C) - 27.2 eV
D) + 27.2 eV
Show Answer
Answer:
Correct Answer: B
Solution:
Energy in nth can be expressed as
$E_n=\frac{E_o}{n^2}$
Second orbit energy $E_2 = −\frac{54.42}{2}$
$E_2 = -13.6 eV$
BETA
