Chapter 26, Problem 41A

To determine

The mass of the other silicon isotope.

Answer to Problem 41A

  $m_2=$ 35.44 protons mass.

Explanation of Solution

Given:

Smaller radius is $r_1=16.23\text{cm}$ .

Larger radius is $r_2=17.97\text{cm}$ .

Mass corresponds to smaller radius is $m_1=$ 28 times protons masses.

Formula used:

The charge to mass ratio of an ion in a mass spectrometer is given by

  $\frac{q}{m}=\frac{2V}{B^2{r}^2}$  ...... (1)

Here, $m,v,q$ and $B$ and $r$ are mass, velocity, charge and magnetic field and radius respectively.

Calculation:

Simplified equation (1) for $m$ ,

  $m=\frac{B^2{r}^{2}q}{2V}$  ...... (2)

Let, the smaller radius $r_1$ corresponds to the mass $m_1$ , therefore,

  $m_1=\frac{B^2{r}_1{}^{2}q}{2V}$  ...... (3)

Let the larger radius $r_2$ corresponds to mass $m_2$ , therefore,

  $m_2=\frac{B^2{r}_2{}^{2}q}{2V}$  ...... (4)

Divide equation (3) by (4)

  $\frac{m_1}{m_2}=\frac{r_1{}^2}{r_2{}^2}$

  $m_2=\frac{m_1{r}_2{}^2}{r_1{}^2}$  ...... (5)

Conversion of cm to m:

  $\begin{array}{l}{r}_1=16.23\text{cm}\\ r_1=\left(16.23\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\\ r_1=0.16\text{m}\end{array}$

And

  $\begin{array}{l}{r}_2=17.97\text{cm}\\ r_2=\left(17.97\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\\ r_2=0.18\text{m}\end{array}$

Substitute $28\left(1.67\times {10}^{-27}\text{kg}\right)$ for $m_1$ , $0.18\text{m}$ for $r_2$ and $0.16\text{m}$ for $r_1$ in equation (5)

  $\begin{array}{l}{m}_2=\frac{\left(28\left(1.67\times {10}^{-27}\text{kg}\right)\right){\left(0.18\text{m}\right)}^2}{{\left(0.16\text{m}\right)}^2}\\ m_2=35.44\left(1.67\times {10}^{-27}\text{kg}\right)\\ \end{array}$

Or

  $m_2=$ 35.44 protons mass.

Conclusion:

Hence, the mass corresponds to larger radius is $m_2=$ 35.44 proton mass.