Discussion Forum

1)

If the electron in hydrogen atom jumps from second Bohr orbit to ground state and difference between 1energies of the two states is radiated in the form of photons. if the work function of the material is 

[ Energy of electron in nth orbit = $-\frac{13.6}{n^{2}}eV$]

Answer:

Option D

Explanation:

 The energy difference between two sates  is given as,

$\triangle E=E_{2}-E_{1}=\frac{-13.6}{2^{2}}-(\frac{-13.6}{1^{2}})$

$\Rightarrow$   $\triangle E=\frac{13.6}{1^{2}}-\frac{13.6}{2^{2}}$

$\Rightarrow$   $\triangle E=13.6\left[\frac{4-1}{4}\right]=13.6 \times\frac{3}{4}$

$\therefore$     $\triangle E$= 10.2 e V

 Since , the energy is radiated  in form of photons, so we have

 Energy of photons =hv =10.2 eV

 From Einstein 's photoelectric equation, we have 

      $hv=\phi_{0}+eV_{s}$

Substituting the given values.

$\therefore$   10.2 eV=4.2e V+e$V_{s}$

$\Rightarrow$     6eV=eV_s

$\therefore$    $V_{s}$= 6 V