Question

What are the energy $\left(kj/mol\right)$ associated with the de-excitation of an electron from $n=6$ to $n=2$ in He ion?

• A. $1.36\times {10}^3$
• B. $1.17\times {10}^{-22}$
• C. $1.16\times {10}^3$
• D. $1.78\times {10}^{-22}$

Answer 1

Answer: (B)

$1.17\times {10}^{-22}$

$\because \frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]\times Z^2$

$\therefore \frac{1}{\lambda }=1.07\times {10}^7[\frac{-1}{36}+\frac{1}{4}]\times 4=0.237\times {10}^7\times 4$

$\therefore \lambda =4.22\times {10}^{-7}\times 4=16.88\times {10}^{-7}$

$E=\frac{hc}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4.22\times {10}^{-7}\times 4}=1.175\times {10}^{-22}KJ/mol$

Hence the correct option is B.