Question

The electron energy in hydrogen atom is given by $E_n=\frac{-21.7\times {10}^{-12}}{n^2}$ergs. The energy required to remove $e^{-}$ completely from n = 2 orbit is:

• A. $5.425\times {10}^{-12}ergs$
• B. $5.467\times {10}^{-12}ergs$
• C. $6.201\times {10}^{-12}ergs$
• D. $6.112\times {10}^{-12}ergs$

Answer 1

Answer: (A)

$5.425\times {10}^{-12}ergs$

Given that,

$E_n=\frac{-21.7\times {10}^{-12}}{n^2}$ergs

For $n=2$
$E_n=\frac{-21.7\times {10}^{-12}}{2^2}$ergs

$\Rightarrow E_n=-5.425\times {10}^{-12}$ergs

Energy used to remove electron is ionisation energy which is equal to the energy used to take the electron to $n=\infty$.

Therefore,
$E=E_{\infty }-E_2$

$\Rightarrow E=0-(-5.425\times {10}^{-12})$

$\Rightarrow E=5.425\times {10}^{-12}$