$(h\, = 6.626 \times 10^{-34}\, Js, N_A\, = 6.022 \times 10^{23}\, mol^{-1} )$
$495.5 = 6.023 \times 10^{23} \times 6.6 \times 10^{-34} \times v$
$v = \frac{{495.5 \times {{10}^3}\,J}}{{6.023 \times {{10}^{23}} \times 6.6 \times {{10}^{ - 34}}}} = 12.4 \times {10^{14}}$
$ = 1.24 \times {10^{15}}\,{s^{ - 1}}$
[આપેલ:પ્લેટીનમની દેહલી આવૃત્તિ $1.3$ $\times 10^{15} \,s ^{-1}$ અને $h =6.6 \times 10^{-34} \,J \,s$.]