Energy of electron in the second orbit of hydrogen atom is E. The energy of electron ' $E_3$ ' in the third…
Energy of electron in the second orbit of hydrogen atom is E. The energy of electron ' $E_3$ ' in the third orbit of helium ( $\left.\mathrm{He}\right)$ atom will be
$E_3=\frac{4 E}{9}$
$E_3=\frac{16 E}{3}$
$E_3=\frac{16 E}{9}$
$E_3=\frac{4 E}{3}$
Solution
The energy for nth orbit in the atom of atomic number \((Z)\) is given by \(E_n=\frac{Z^2}{n^2} E_0\) So, for helium \((Z=2)\) and \(\mathrm{n}=3\)
\(\therefore\) for third orbit \(E_3=\frac{4}{9} E_0\) ?(1)
Now for H atom \((\mathrm{Z}=1)\) and \(\mathrm{n}=2\)
\(\therefore\) for second orbit \(E_2=\frac{1}{4} E_0\) ?(2)
Now from Eqs. (1) and (2),
we obtain \(\frac{E_3}{E_2}=\frac{\frac{4}{9} \varepsilon_0}{\frac{1}{4} E_0}=\frac{16}{9}\) or \(E_3=\frac{16}{9} E_2=\frac{16}{9} E\)