Chapter 27, Problem 96P

(a)

To determine

The work function of the metal.

Answer to Problem 96P

The work function of the metal is $\underset{\_}{4.7\text{eV}}$.

Explanation of Solution

Given that the wavelength of light is $180\text{nm}$, the magnetic field strength is $7.5\times {10}^{-5}\text{T}$, and the radius of the circular path is $6.7\text{cm}$

Write the expression for the magnitude of the magnetic force on the electron when it enters perpendicular into the field.

  $F=evB$                                                                                                                  (I)

Here, $F$ is the magnetic force, $e$ is the charge of electron, $v$ is the speed of electron, and $B$ is the magnetic field strength.

The magnetic force is providing the necessary centripetal force for the motion of the electron. Thus, the magnetic force can be equated to the centripetal force on electron.

  $evB=\frac{m{v}^2}{r}$                                                                                                           (II)

Here, $m$ is the mass of electron, and $r$ is the radius of the circular path.

Solve equation (II) for $v$.

  $v=\frac{eBr}{m}$                                                                                                               (III)

Write the expression for the maximum kinetic energy of the ejected electron.

  $K_{\text{max}}=\frac{1}{2}m{v}^2$                                                                                                      (IV)

Here, $K_{\text{max}}$ is the maximum kinetic energy of the ejected electron.

Write the expression for the maximum kinetic energy of the ejected electron in terms of the work function of the metal.

  $K_{\text{max}}=\frac{hc}{\lambda }-\varphi$                                                                                                    (V)

Here, $c$ is the speed of light,$\lambda$ is the  $h$ is the Planck’s constant, and $\varphi$ is the work function of the metal.

Equate the right hand sides of equations (IV) and (V) and solve for $\varphi$.

  $\varphi =\frac{hc}{\lambda }-\frac{1}{2}m{v}^2$                                                                                                     (VI)

Use equation (III) in (VI).

  $\begin{array}{c}\varphi =\frac{hc}{\lambda }-\frac{1}{2}m{\left(\frac{eBr}{m}\right)}^2\\ =\frac{hc}{\lambda }-\frac{{\left(eBr\right)}^2}{2m}\end{array}$                                                                                           (VI)

Conclusion:

Substitute $180\text{nm}$ for $\lambda$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3.00\times {10}^8\text{m/s}$ for $c$, $1.602\times {10}^{-19}\text{C}$ for $e$, $7.5\times {10}^{-5}\text{T}$ for $B$, $6.7\text{cm}$ for $r$, and $9.1\times {10}^{-31}\text{kg}$ for $m$ in equation (VI) to find $\varphi$.

  $\begin{array}{c}\varphi =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{m/s}\right)}{180\text{nm}}-\frac{{\left[\left(1.602\times {10}^{-19}\text{C}\right)\left(7.5\times {10}^{-5}\text{T}\right)\left(6.7\text{cm}\right)\right]}^2}{2\left(9.1\times {10}^{-31}\text{kg}\right)}\\ =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{m/s}\right)}{180\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{nm}}}-\frac{{\left[\left(1.602\times {10}^{-19}\text{C}\right)\left(7.5\times {10}^{-5}\text{T}\right)\left(6.7\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right]}^2}{2\left(9.1\times {10}^{-31}\text{kg}\right)}\\ =7.4833\times {10}^{-19}\text{J}\times \frac{1\text{eV}}{1.602\times {10}^{-19}\text{J}}\\ =4.7\text{eV}\end{array}$

Therefore, the work function of the metal is $\underset{\_}{4.7\text{eV}}$.

(b)

To determine

Whether the electrons with maximum kinetic energy follow a path with the maximum or minimum radius.

Answer to Problem 96P

The electrons with maximum kinetic energy follow a path with the maximum radius.

Explanation of Solution

Equation (IV) indicates that the maximum kinetic energy of the electron is directly proportional to the square of the speed of the electrons.

  $K\propto v^2$

Equation (III) indicates that the speed of the electron is directly proportional to the radius of the circular path.

  $v\propto r$

Hence, it can be summarized that the maximum kinetic energy of the electron corresponds to the large value of the radius. Thus, The electrons with maximum kinetic energy follow a path with the maximum radius.

Conclusion:

Therefore, the electrons with maximum kinetic energy follow a path with the maximum radius.