Chapter 19, Problem 34P

(a)

To determine

What is the mass of the unknown element?

Answer to Problem 34P

Mass of the unknown element is $39\text{u}$.

Explanation of Solution

Sulfur, manganese and the unknown element are accelerated through the same potential difference. Also they have same charge and kinetic energy. Therefore the equations for kinetic energy of sulfur and unknown element can be equated.

$\frac{1}{2}{m}_s{v}_s{}^2=\frac{1}{2}{m}_u{v}_u{}^2$ (I)

Here, $m_s$ is the mass of sulfur ion, $v_s$ is the speed of sulfur ion, $m_u$ is the mass of unknown element, $v_u$ is the speed of unknown element

Simplify equation (I) to get the ratio of $v_s\text{and }{v}_u$

$\frac{v_s}{v_u}=\sqrt{\frac{m_u}{m_s}}$ (II)

Equate the equations for centripetal acceleration and acceleration due to the magnetic field

$\frac{v^2}{r}=\frac{qvB}{m}$ (III)

Here, $v$ is the velocity of ion, $r$ is the radius of the circular path , $q$ is the charge of the ion, $B$ is the magnetic field, $m$ is the mass of the ion

Simplify equation (III) to get $r$

$r=\frac{mv}{qB}$ (IV)

Rewrite equation (IV) to get $r_s$ and $r_u$.

$r_s=\frac{m_s{v}_s}{qB}$ (V)

Here, $r_s$ is the radius of circular path travelled by sulfur ion, $m_s$ is the mass of the sulfur ion, $v_s$ is the speed of sulfur ion, $q$ is the charge, $B$ is the magnetic field

$r_u=\frac{m_u{v}_u}{qB}$ (VI)

Here, $r_u$ is the radius of path of unknown ion, $m_u$ is the mass of unknown ion, $v_u$ is the speed of unknown ion, $q$ is the charge of ion, $B$ is the magnetic field

Take ratio of $r_s$ and $r_u$ from equations (V) and (VI).

$\frac{r_s}{r_u}=\frac{m_s{v}_s}{m_u{v}_u}$ (VII)

Substitute for $\frac{v_s}{v_u}$ from equation (II) to equation (VII) and simplify for $m_u$

$\begin{array}{c}\frac{r_s}{r_u}=\frac{m_s}{m_u}\sqrt{\frac{m_u}{m_s}}\\ =\sqrt{\frac{m_s}{m_u}}\\ m_u={\left(\frac{r_u}{r_s}\right)}^2{m}_s\\ ={\left(\frac{r_s+r_u}{r_s}\right)}^2{m}_s\end{array}$ (VIII)

Similarly doing find the ratio of radius between manganese ion and sulfur ion.

$\frac{r_{Mn}}{r_s}=\sqrt{\frac{m_{Mn}}{m_s}}$ (IX)

Simplify equation (IX) to get $r_{Mn}$

$r_{Mn}=r_s\sqrt{\frac{m_{Mn}}{m_s}}$ (X)

Simplify equation (X)

$\begin{array}{c}{r}_{Mn}-r_s=r_s\left(\sqrt{\frac{m_{Mn}}{m_s}-1}\right)\\ r_s=\frac{r_{Mn}-r_s}{\sqrt{\frac{m_{Mn}}{m_s}}-1}\end{array}$ (XI)

Conclusion

Substitute $1.07\text{cm}$ for $r_u$ , in equation (X) to get $m_u$

$\begin{array}{c}{m}_u={\left(\frac{r_s+1.07\text{cm}}{r_s}\right)}^2{m}_s\\ ={\left(1+\frac{1.07\text{cm}}{r_s}\right)}^2{m}_s\end{array}$

Substitute $3.20\text{cm}$ for $r_{Mn}-r_s$ , $55\text{u}$ for $m_{Mn}$ , and $32\text{u}$ for $m_s$ in equation (XI) to get $r_s$

$\begin{array}{c}{r}_s=\frac{3.20\text{cm}}{\sqrt{\frac{55\text{u}}{32\text{u}}}-1}\\ =10.3\text{cm}\end{array}$

Substitute $10.3\text{cm}$ for $r_s$ in equation for $m_u$

$\begin{array}{c}{m}_u={\left(1+\frac{1.07\text{cm}}{10.3\text{cm}}\right)}^2\left(32\text{u}\right)\\ =39\text{u}\end{array}$

Therefore, the mass of the unknown element is $39\text{u}$

(b)

To determine

Which is the unknown element?

Answer to Problem 34P

The unknown element is potassium.

Explanation of Solution

Mass of the unknown element is found as $39\text{u}$. Clearly from the periodic table, the element having mass number $39\text{u}$ is identified as potassium.

Conclusion:

Therefore, The unknown element is potassium.