If the first excitation potential of a hypothetical hydrogen like atom is \(15 \mathrm{~V}\), then the third…
If the first excitation potential of a hypothetical hydrogen like atom is \(15 \mathrm{~V}\), then the third excitation potential of the atom is
\(13.6 \mathrm{~V}\)
\(\frac{4}{75} \mathrm{~V}\)
\(\frac{15}{16} \mathrm{~V}\)
\(\frac{75}{4} \mathrm{~V}\)
Solution
Let total energy of hydrogen like atom at ground state \(=-E\)
So, Ist excitation
\(\begin{aligned}
\text {Energy } & =E_{2 n d}-E_{1 s t} \\
15 \mathrm{eV} & =\left(\frac{-E}{2^2}\right)-\left[\frac{E}{1^2}\right]
\end{aligned}\)
( \(\because\) First excitation potential is given)
\(\begin{array}{ll}
\Rightarrow & \frac{3 E}{4}=15 \mathrm{eV} \\
\Rightarrow & E=\frac{15 \times 4}{3} \mathrm{eV}=20 \mathrm{eV}
\end{array}\)
Now, the third excitation energy
\(\begin{aligned}
E_3 & =E_{4^{\text {th }}}-E_{1^{\mathrm{st}}} \\
& =\left(\frac{-E}{4^2}\right)-\left[\frac{E}{1^2}\right] \\
E_3 & =+\frac{15 E}{16}
\end{aligned}\)
Now, putting the value of \(E\), we get
\(E_3=\frac{15}{16} \times 20=\frac{75}{4} \mathrm{eV}\)
Hence, the correct option is (d).