Question
The ionization energy of a hydrogen-like Bohr atom is $$4$$ Rydbergs. Find the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state:
$$[1\,Rydberg = 2.2 \times {10^{ - 18}},h = 6.6 \times {10^{ - 34}}Js,c = 3 \times {10^8}m/s.$$ Bohr radius of hydrogen atom $$ = 5 \times {10^{ - 11}}m$$ ]
A.
$$400\,\mathop {\text{A}}\limits^ \circ $$
B.
$$300\,\mathop {\text{A}}\limits^ \circ $$
C.
$$500\,\mathop {\text{A}}\limits^ \circ $$
D.
$$600\,\mathop {\text{A}}\limits^ \circ $$
Answer :
$$300\,\mathop {\text{A}}\limits^ \circ $$
Solution :
The energy in the ground state
$$\eqalign{
& {E_1} = - 4\,Rydberg \cr
& {E_1} = - 4 \times 2.2 \times {10^{ - 18}}J \cr} $$
The energy of the first excited state $$\left( {n = 2} \right)$$
$${E_2} = \frac{{{E_1}}}{4} = - 2.2 \times {10^{ - 18}}J$$
The energy difference
$$\Delta E = {E_2} - {E_1} = 3 \times 2.2 \times {10^{ - 18}}J$$
Now, the wavelength of radiation emitted is
$$\eqalign{
& \lambda = \frac{{hc}}{{\Delta E}} \cr
& \lambda = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3 \times 2.2 \times {{10}^{ - 18}}}} = 300\,\mathop {\text{A}}\limits^ \circ \cr} $$