Question

Light of 300 nm wavelength strikes a metal plate, and photoelectrons are produced moving as fast as 0.002 c.(a) What is the work function of the metal? (b) What is the threshold wavelength for this metal?

Short answer

(a) The work function of the metal is 3.11 eV .

(b)The threshold wavelength of the metal is 398.7 nm .

Step-by-step solution

Identification of given data

The given data is listed as follows,

The wavelength of the light is, $\mathrm{\lambda }=300\mathrm{nm}\times \left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=3.00\mathrm{x}{10}^{-7}\mathrm{m}$

The speed of photo electron is, v = 0.002c

Expression of the kinetic energy and Planck’s formula

The expression for the kinetic energy of electrons is given as follows,

$\mathbf{KE}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Here, m is the mass of electrons, and v is the velocity of electrons.

The expression for the Planck’s formula is given as follows,

$\mathbf{\varphi }\mathbf{=}\frac{\mathbf{hc}}{\mathbf{\lambda }}$

Here, h is the Planck’s constant, c is the speed of light, and $\mathit{\lambda }$ is the wavelength.

(a) Determination of work function of the metal

Write the expression for the maximum kinetic energy depending on the Planck’s constant.

$K{E}_{max}=\frac{hc}{\lambda }-\varphi$

Substitute the value of kinetic energy and rearrange.

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\varphi \\ \varphi =\frac{hc}{\lambda }-\frac{1}{2}m{v}^2 \end{gathered}$$

It is known that hc = 1240 eV.nm ,and mass of electron is $9.1\times {10}^{-31}\mathrm{kg}$. Substitute these values in the above expression.

$$\begin{gathered} \mathrm{\varphi }=\frac{1240\mathrm{eV}·\mathrm{nm}}{300\mathrm{nm}}-\frac{9.1\times {10}^{-31}\mathrm{kg}\times {\left(0.002\times \mathrm{c}\right)}^2}{2} \\ =\frac{1240\mathrm{eV}·\mathrm{nm}}{300\mathrm{nm}}-\frac{9.1\times {10}^{-31}\mathrm{kg}\times {\left(0.002\times 3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}{2} \\ =4.13\mathrm{eV}-1.638\times {10}^{-19}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}\times \frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}} \\ =\left(4.13\mathrm{eV}-1.023\mathrm{eV}\right) \\ =3.11\mathrm{eV} \end{gathered}$$

Thus, the work function of the metal is 3.11 eV .

(b) Determination of threshold wavelength of the metal

The work function is same as that of the energy of the upcoming photon$K{E}_{max}=\frac{hc}{\lambda }-\varphi$and it is known that the ejected electron is having zero kinetic energy, that is$K{E}_{max}=0$.

Apply the above condition and write the expression.

$$\begin{gathered} \frac{hc}{\lambda }-\varphi =0 \\ \frac{hc}{\lambda }=0 \\ \lambda =\frac{hc}{\varphi } \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} \mathrm{\lambda }=\frac{1240\mathrm{eV}·\mathrm{nm}}{3.11\mathrm{eV}} \\ =398.7\mathrm{nm} \end{gathered}$$

Thus, the threshold wavelength of the metal is 398.7 nm .