Question

An H-like ion is observed to emit six different wavelengths originating from all possible transitions between a group of levels. These levels have energies between $-085$ eV (Min) and $-0.544$ eV (Max).
(a) Find the atomic number (z) of an element
(b) Find quantum numbers of levels between which transitions occur.
(c) Calculate the largest wavelength emitted in transition between the levels.

Answer 1

Do small correction in this question

Maximum energy should $6e-5.44eV$

Minimum energy $=0.85eV$.

Let minimum level be $n_1$

Do maximum level should be $n_1+2$

$\because$ Total number of transition is given $6$

For maximum energy

$-5.44=-13.6(\frac{1}{n_1^2}-\frac{1}{(n_1+2{)}^2})z^2$

$5.44=13.6(\frac{1}{n_1^2}-\frac{1}{(n_1+2{)}^2})z^2$ ….(i)

For minimum energy

$-0.85=-13.6(\frac{1}{(n_1+1{)}^2}-\frac{1}{(n_1+2{)}^2})z^2$ ……(ii)

Divide equation (i) by (ii)

$\frac{5.44}{0.85}=\frac{(\frac{1}{n_1^2}-\frac{1}{(n_1+2{)}^2})z^2}{(\frac{1}{(n_1+1{)}^2}-\frac{1}{(n_1+2{)}^2})z^2}$

$\frac{32}{5}=\frac{\left(\right(n_1+2{)}^2-n_1^2)(n_1+1{)}^2}{\left[\right(n_1+2{)}^2-(n_1+1{)}^2]n_1^2}$

$\frac{32}{5}=\frac{(n_1^2+4n_1+4-n_1^2)(n_1+1{)}^2}{(n_1^2+4n_1+4-n_1^{2}2n_1-1)n_1^2}$

$\frac{32}{5}=\frac{4(n_1+1{)}^3}{(2n_1+3)n_1^2}$

$8(2n_1^2+3n_1^2)=5(n_1+1{)}^3$

$16n_1^3+24n_1^2=3n_1^3+15n_1^2+15n_1+5$

$11n_1^3+9n_1^2-15n_1-5=0$

$(n_1-1)(11n_1^2+20n_1+5)=0$

$n_1=1$

Put the value of $n_1$ in equation (i)

$5.44=13.6(1-\frac{1}{3^2})z^2$

$5.44=13.6\times \frac{{82}^2}{2}$

$\frac{5.44\times 9}{13.6\times 8}=z^2$

$z^2=9$

$z=3$

(a) $z=3$

(b) $n_1=1,n_2=3$

(c) $\frac{1}{\lambda }=R_H(1-\frac{1}{2^2})\Rightarrow \lambda =\frac{4}{3R_H}=121.5$ nm.