Chapter 7, Problem 7.45QP

Interpretation Introduction

Interpretation: The thin layers of potassium and sodium are given to be exposed to radiation of wavelength $300\text{nm}$. The metal that emits electrons with the greater velocity is to be identified. Also the velocity of these electrons is to be calculated.

Concept introduction: Potassium and Sodium are the elements that are present in the periodic table. The velocity of an object tells the rate at which the object is moving with respect to time and position.

To determine: The velocity of the given electrons and the metal that emits electrons with the greater velocity.

Answer to Problem 7.45QP

Solution

The velocity of electrons ejected by potassium is $\underset{\_}{8.04\times 10^{5}m/s}$.

The velocity of electrons ejected by sodium is $\underset{\_}{6.97\times 10^{5}m/s}$.

Thus, the potassium metal emits the electron with greater velocity.

Explanation of Solution

Explanation

Given

The radiation of wavelength, $\lambda$ exposed to potassium and sodium is $300\text{nm}$.

Work function for thin layer potassium, $\Phi$ is $3.68\times {10}^{-19}\text{J}$.

Work function for thin layer sodium, $\Phi$ is $4.41\times {10}^{-19}\text{J}$.

The energy of a photon is calculated by the formula,

$\begin{array}{c}E=\frac{hc}{\lambda }\\ E=h{v}_{°}\\ =\Phi \end{array}$ (1)

Where,

• $\Phi$ is the work function for elements.
• $E$ is the energy of a photon.
• $v_o$ is the threshold frequency.
• $h$ is the Planck’s constant $\left(6.62\times {10}^{-34}\text{Js}\right)$.
• $c$ is the velocity of light $\left(3\times {10}^8\text{m}/\text{s}\right)$.

The wavelength in $\text{nm}$ is converted into $\text{m}$ given as,

$300\text{nm}=300\times {10}^{-9}\text{m}$

Substitute the value of wavelength, wavelength and velocity of light in equation (1).

$\begin{array}{c}E=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{300\times {10}^{-9}}\\ =6.62\times {10}^{-19}\text{J}\end{array}$

As, $E=hv$ therefore, $hv=6.62\times {10}^{-19}\text{J}$

The work function for potassium, $\Phi =h{v}_0$ is $3.68\times {10}^{-19}\text{J}$.

The kinetic energy of an electron is calculated by the formula,

$KE=hv-h{v}_o$ (2)

Substitute the value of $hv$ and $h{v}_o$ in the equation (2).

$\begin{array}{c}KE=\left(6.62\times {10}^{-19}-3.68\times {10}^{-19}\right)\text{J}\\ \text{=2}\text{.94}\times {\text{10}}^{-19}\text{J}\end{array}$

The velocity is calculated by the formula,

$v=\sqrt{\frac{2KE}{M}}$ (3)

Substitute the value of Kinetic energy and mass of an electron in equation (3).

$\begin{array}{c}v=\sqrt{\frac{2\times 2.94\times {10}^{-19}}{9.11\times {10}^{-31}}}\\ =\sqrt{\frac{5.88\times {10}^{-19}}{9.11\times {10}^{-31}}}\\ =\sqrt{0.5036\times {10}^{12}}\\ =\underset{\_}{8.04\times 10^{5}m/s}\end{array}$ (4)

Therefore, the velocity of electron emitted by potassium is $\underset{\_}{8.04\times 10^{5}m/s}$.

The work function for sodium, $\Phi =h{v}_0$ is $4.41\times {10}^{-19}\text{J}$.

The kinetic energy of an electron is calculated by the formula,

$KE=hv-h{v}_o$ (5)

Substitute the value of $hv$ and $h{v}_o$ in the equation (5).

$\begin{array}{c}KE=\left(6.62\times {10}^{-19}-4.41\times {10}^{-19}\right)\text{J}\\ \text{=2}\text{.21}\times {\text{10}}^{-19}\text{J}\end{array}$

The velocity is calculated by the formula,

$v=\sqrt{\frac{2KE}{M}}$ (6)

Substitute the value of Kinetic energy and mass of an electron in equation (6).

$\begin{array}{c}v=\sqrt{\frac{2\times 2.21\times {10}^{-19}}{9.11\times {10}^{-31}}}\\ =\sqrt{\frac{4.42\times {10}^{-19}}{9.11\times {10}^{-31}}}\\ =\sqrt{0.485\times {10}^{12}}\\ =\underset{\_}{6.97\times 10^{5}m/s}\end{array}$ (7)

Thus, the velocity of electron emitted by sodium is $\underset{\_}{6.97\times 10^{5}m/s}$.

Therefore, on comparing the equation (4) and equation (7), it is concluded that electrons ejected from potassium element have greater velocity.

Conclusion

The velocity of electron emitted by potassium is $\underset{\_}{8.04\times 10^{5}m/s}$.

The velocity of electron emitted by sodium is $\underset{\_}{6.97\times 10^{5}m/s}$.

Thus, the velocity of electrons will be more in the case of potassium element.