Question

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

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

Answer 1

Answer: (A)

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