$\therefore \,B{E_{(C - H)}} = 90{\kern 1pt} kJ/mol$

In ${C_2}{H_6},\,B{E_{(C - C)}} + 6 \times B{E_{(C - H)}} = 620\,kJ/mol$

$\therefore \,B{E_{(C - C)}} = 80{\kern 1pt} kJ/mol$

$\therefore \,B{E_{(C - C)}} = \frac{{80 \times {{10}^3}}}{{6.023 \times {{10}^{23}}}}\,J/molecule$

Now, $E = \frac{{hc}}{\lambda }$

$\therefore \,\lambda  = \frac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 6.023 \times {{10}^{23}}}}{{80 \times {{10}^3}}}$

$\therefore \,\lambda  = 1.419 \times {10^3}\,nm$