Question

Calculate the wave number
Of Stretching Vibration of
a carbon carbon double
bond. Give force constant
(k = l0×10^5 dynes cm-1)​

Answer 1

Answer:see on picture

Explanation:

Answer 2

Given: stretching of C=C double bond. Force constant of $k=10*10^{5} dynes/cm$.

To find: the wave number of stretching vibration C=C.

Step-by-step method:

Step 1 of 2

The formula of wave number is:

1/λ$=\frac{1}{2\pi c} \sqrt{\frac{K}{M} }$

The reduced mass, μ =$\frac{m_{1}m_{2} }{m_{1}+m_{2}}$

$=\frac{12*12}{12+12} \\=\frac{144}{24} \\=6$

Step 2 of 2

1/λ$=\frac{1}{2\pi c} \sqrt{\frac{K}{M} }$

$=1307\sqrt{\frac{10 }{6} }\\=1307 *1.29\\=1,687.32$

The wave number is $1,687.32cm^{-1}$