problem 36 and 37 The 12C160 molecule has an equilibrium bond distance of 112.8
ID: 771344 • Letter: P
Question
problem 36 and 37
Explanation / Answer
a) reduced mass = m1m2/(m1+m2 ) = (12x16)/(12+16) = 6.857 amu , = 6.857 x1.66 x10^ -27 = 1.138 x10^ -26 kg b) I = mr^2 = 1.138 x10^ -26 x (112.8 x10^ -12)^2 = 1.448 x10^ -46 kgm^2 , 37) a) lamda x T = 0.002897 , is formula , lamda = 600 nm = 600 x10^ -9 m , T = 0.002897 /(600 x10^ -9) = 4828.33 K is temp of star , b) T = 0.002897/( 800 x 10^ -9) = 3621.25 K is temp of rod ,
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