TG EAMCET 3 May 2025 Shift 2 Paper

Step 1: Find the Total Energy (TE) from the Given Potential Energy (PE)The potential energy of the electron is given as -6.8 eV .We use the relation: $PE = 2(TE)$This means: $TE = \frac{PE}{2} = \frac{-6.8}{2} = -3.4 \text{ eV}$Step 2: Find the Value of $n$ (Principal Quantum Number)For a hydrogen atom, the total energy formula is: $TE = \frac{-13.6}{n^2}$Set this equal to the TE we found: $-3.4 = \frac{-13.6}{n^2}$Solve for $n^2: n^2 = 4$So, $n=2$Step 3: The Electron is in the Second OrbitSince $n=2$, the electron is in the second Bohr orbit.Step 4: Use Bohr's Model and de-Broglie HypothesisAccording to Bohr and de-Broglie:$2 \pi r_n = n \lambda$This gives: $\lambda = \frac{2 \pi r_n}{n}$Step 5: Use the Formula for the Radius of Orbit Radius of the nth orbit: $r_n = n^2 r_0$For $n=2$ (second orbit): $r_2 = 2^2 r_0 = 4 r_0$Step 6: Calculate the de-Broglie WavelengthPlug $r_2$ into the earlier formula:$\lambda = \frac{2 \pi r_2}{2} = \frac{2 \pi}{2} \times 4 r_0 = 4 \pi r_0$