Chapter 27, Problem 25P

To determine

(a) To determine:

Maximum kinetic energy of the ejected electrons.

Answer to Problem 25P

Solution:

Maximum kinetic energy of the ejected electrons is $0.78\text{eV}$.

Explanation of Solution

Given:

Threshold wavelength, $\lambda =340\text{nm}$

Given wavelength of light, $280\text{nm}$

Formula used:

The maximum kinetic energy of the photon is calculated as

$KE=\frac{hc}{e}\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_o}\right]$

Here,

$e$ is the charge of electron

$h=6.62\times {10}^{-34}\text{J-s}$ is the Planck’s constant

$c=3\times {10}^8\text{m/s}$ is the speed of light

$\lambda$ is the given wavelength

${\lambda }_o$ is the threshold wavelength

Calculation:

The maximum kinetic energy of the photon is calculated as

$KE=\frac{hc}{e}\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_o}\right]$

Plugging the values in the above equation

$\begin{array}{c}KE=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}\left[\frac{1}{280\times {10}^{-9}}-\frac{1}{340\times {10}^{-9}}\right]\\ =0.78\text{eV}\end{array}$

To determine

(b) To determine:

Maximum kinetic energy of the ejected electrons

Answer to Problem 25P

Solution:

Maximum kinetic energy of the ejected electrons is $0$.

Explanation of Solution

Given:

Threshold wavelength, $\lambda =340\text{nm}$

Given wavelength of light, $360\text{nm}$

Formula used:

The maximum kinetic energy of the photon is calculated as

$KE=\frac{hc}{e}\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_o}\right]$

Here,

$e$ is the charge of electron

$h=6.62\times {10}^{-34}\text{J-s}$ is the Planck’s constant

$c=3\times {10}^8\text{m/s}$ is the speed of light

$\lambda$ is the given wavelength

${\lambda }_o$ is the threshold wavelength

Calculation:

The maximum kinetic energy of the photon is calculated as

$KE=\frac{hc}{e}\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_o}\right]$

Plugging the values in the above equation

$\begin{array}{c}KE=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}\left[\frac{1}{360\times {10}^{-9}}-\frac{1}{340\times {10}^{-9}}\right]\\ =-0.2\text{eV}\end{array}$

The negative sign indicates that, the photon will not have the enough energy to eject the electron from the metal surface; therefore, no electron will be ejected from the surface of metal,