Discussion Forum

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number n_i to another with quantum number n_f . v_i and v_f  are respectively, the initial and final potential energies of the electron. If  $\frac{v_{i}}{v_{f}}=6.25$    then the  smallest possible n_f  is

Answer:

Explanation:

Potential energy of hydrogen atom (Z=1) in nth orbit (in eV)

   $P.E=-\frac{27.2}{n^{2}}$

            = $\frac{v_{f}}{v_{i}}=\frac{-\frac{27.2}{n_f^2}}{-\frac{27.2}{n_i^2}}$ =  $\frac{1}{6.25}$

    $6.25=\frac{n_f^2}{n_i^2}$

$\frac{n_{f}}{n_{i}}=2.5=\frac{5}{2}$

 hence the answer is 5