Question

What is the frequency and wave-length of a photon emitted during a transition from the n = 5 state to the n = 2 state in the hydrogen atom.

Answer 1

It's as mentioned in attachment

Answer 2

$\Delta E\quad =\quad 2.18\quad \times \quad { 10 }^{ -18 }\quad \left( \frac { 1 }{ { n }1^{ 2 } } \quad -\quad { n2 }^{ 2 } \right)$

Where $n_1$ = 2 and $n_2$= 5; this transition gives rise to a spectral line in the "Balmer series".

                 $\Delta E\quad =\quad 2.18\quad \times \quad { 10 }^{ -18 }\quad \left( \frac { 1 }{ { 2 }^{ 2 } } \quad -\quad \frac { 1 }{ { 5 }^{ 2 } } \right)$

                 $=\quad 2.18\quad \times \quad { 10 }^{ -18 }\quad \left( \frac { 1 }{ 4 } \quad -\quad \frac { 1 }{ 25 } \right)$

                 \Delta E\quad =\quad 4.58\quad \times \quad { 10 }^{ -19 }\quad J

We know, $\Delta E\quad =\quad hv$

$v\quad =\quad \frac { 4.58\quad \times \quad { 10 }^{ -19 }\quad J }{ 6.62\quad \times \quad { 10 }^{ -34 }\quad Js }$

Where, h = Planck's constant $= 6.62 \times { 10 }^{ -34 } Js$

                  $v\quad =\quad 6.91 \times { 10 }^{ 41 } { s }^{ -1 }$

Also, $\lambda = \frac { c }{ v } = \frac { 3.0 \times { 10 }^{ 8 } m/s }{ 6.91 \times  { 10 }^{ 14 } { s }^{ -1 } }$

                  $\lambda = 434 \times { 10 }^{ -9 } m\\\lambda = 434 nm$