Question

IE for $He ^{+}$is $1.96 \times 10^{-19} J$ atom ${ }^{-1}$. Calculate the energy of first stationary state of $Be ^{+3}$

• A. $7.84 \times 10^{-19} J$ atom $^{-1}$
• B. $7.84 \times 10^{-23} J $ atom $^{-1}$
• C. $7.84 \times 1019 J $ atom $^{-1}$
• D. $7.84 \times 1023 J$ atom $^{-1}$

Answer 1

Answer: (A)

For $He ^{+}, E_1=E_1$ for $H \times Z^2=E_1 \times 4$
For $Be ^{3+}, E_1=E_1$ for $H \times Z^2=E_1 \times 16$
For $Be ^{+3}, E_1=E_1$ for $He ^{+} \times \frac{16}{4}$
$=1.96 \times 10^{-19} \times 4$
$=7.84 \times 10^{-19} J /$ atom