JEE Main 2014 · 9 Apr · Shift 1 · Chemistry Q38

ChemistrySingle correct+4 / -1Question 38 of 90

The energy of an electron in first Bohr orbit of H - atom is $-13.6$ eV. The energy value of electron in the excited state of $\ce{Li^2+}$ is :

-27.2

30.6 eV

-30.6

27.2 eV

Answer and worked solution

Official answer: C

$E_n=-13.6\dfrac{Z^2}{n^2}$ eV. For $\ce{Li^2+}$ in its first excited state ( $n=2$ ): $-13.6\times\frac94=-30.6$ eV.

JEE Main 2014 · 9 Apr · Shift 1, Chemistry question 38