Question

The energy of an electron in the first Bohr orbit of hydrogen atom is \(-13.6\,eV\). The energy of the electron in the third orbit is

• A. \(-1.51\,eV\)
• B. \(-3.4\,eV\)
• C. \(-13.6\,eV\)
• D. \(-0.85\,eV\)

Hint:

Bohr Energy Formula: \[ \boxed{E_n=\frac{-13.6}{n^2}\;eV} \] For hydrogen atom: \[ E_1=-13.6\,eV \] \[ E_2=-3.4\,eV \] \[ E_3=-1.51\,eV \] Frequently asked CUET numerical.

Answer 1

The Correct Option is A

Step 1: Recall Bohr's energy formula. \[ E_n=\frac{-13.6}{n^2}\;eV \] For the third orbit: n=3

Step 2: Substitute the value of \(n\). \[ E_3=\frac{-13.6}{3^2} \] \[ E_3=\frac{-13.6}{9} \] \[ E_3=-1.51\,eV \]

Step 3: Choose the correct option. \[ \boxed{E_3=-1.51\,eV} \] Therefore, \[ \boxed{\text{(A)}} \] is the correct answer.