Chapter 9, Problem 9.35E

Determine the speed of an electron being emitted by rubidium $\left(\varphi =\text{2}\text{.16}\text{eV}\right)$ when light of the following wavelengths is shined on the metal in vacuum: (a) $\text{550}\text{nm,}$ (b) $\text{450}\text{nm,}$(c) $\text{350}\text{nm}$.

Interpretation Introduction

Interpretation:

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is to be stated.

Concept introduction:

The emission of electrons when the light is shone upon a material is called the photoelectric effect. The electrons emitted by such a method are also known as photo-electrons. The process basically involves shining light of frequency $\upsilon$, on a metal surface to remove the electrons with a kinetic energy of $\frac{\text{1}}{\text{2}}\text{m}{v}^{\text{2}}$.

Answer to Problem 9.35E

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is $\text{1}{\text{.8×10}}^{\text{5}}{\text{msec}}^{-\text{1}}$.

Explanation of Solution

The given value of work function $\varphi$ is $\text{2}\text{.16 eV}$ and $\lambda$ is $\text{550nm}$.

The formula to calculate the velocity of an electron is given below.

$\begin{array}{rll}\frac{\text{hc}}{\lambda }& =& \varphi +\frac{1}{2}\text{m}{v}^2\cdot \cdot \cdot (1)\\ \end{array}$

Where,

• $\text{h}$ is the Planck’s Constant $\left(\text{6}{\text{.6×10}}^{-\text{34}}\text{ J sec}\right)$

• $\text{c}$ is the speed of light $\left(3.8\times {10}^8{\text{ m sec}}^{-\text{1}}\right)$

• $\lambda$ is the wavelength

• $\text{m}$ is the mass of an electron

• $v$ is the velocity of the emitted electron

• $\varphi$ is the work function

It is known that, $1\text{eV}=\text{1}{\text{.6×10}}^{-\text{31}}\text{J}$

Therefore,

$\text{2}\text{.16}\text{eV×}\frac{\text{1}{\text{.6×10}}^{-\text{31}}\text{J}}{\text{1eV}}=3.456\times {10}^{-\text{19}}\text{J}$

It is known that,

1nm ${=\text{10}}^{-\text{9}}\text{m}$

Therefore,

$\frac{{\text{550nm×10}}^{-\text{9}}}{\text{1nm}}={\text{550×10}}^{-\text{9}}\text{nm}$

Rearrange equation (1) to calculate the value of velocity.

$v=\sqrt{\frac{\text{2}}{\text{m}}\left(\frac{\text{hc}}{\lambda }-\varphi \right)}$

Substitute the values of work function $\varphi$, Planck’s constant $\text{h}$, speed of light $\text{c}$, wavelength $\lambda$ and mass of electron $\text{m}$ in the above equation.

$\begin{array}{rll}v& =& \sqrt{\frac{\text{2}\left(\frac{\text{6}{\text{.6×10}}^{-\text{34}}{\text{Jsec × 3×10}}^{\text{8}}{\text{msec}}^{-\text{1}}}{{\text{550×10}}^{-\text{9}}\text{m}}-\text{3}{\text{.456×10}}^{-\text{19}}\text{J}\right)}{\text{9}{\text{.1×10}}^{-\text{31}}\text{kg}}}\\ v& =& \sqrt{\frac{\text{2}\left(\text{0}{\text{.036×10}}^{-\text{17}}\text{J}-\text{0}{\text{.03456×10}}^{-\text{17}}\text{J}\right)}{\text{9}{\text{.1×10}}^{-\text{31}}\text{kg}}}\\ v& =& \sqrt{\text{3}{\text{.1536×10}}^{\text{10}}{\text{m}}^{\text{2}}{\text{sec}}^{-\text{2}}}\\ v& =& \text{1}{\text{.7758×10}}^{\text{5}}{\text{msec}}^{-\text{1}}\simeq \text{1}{\text{.8×10}}^{\text{5}}\end{array}$

Therefore, the value of speed is $1.8\times {10}^5{\text{ m sec}}^{-\text{1}}$.

Conclusion

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is $\text{1}{\text{.8×10}}^{\text{5}}{\text{ m sec}}^{-\text{1}}$.

Interpretation Introduction

Interpretation:

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is to be stated.

Concept introduction:

The emission of electrons or other free electrons when the light is shone upon a material is called the photoelectric effect. The electrons emitted by such a method are also known as photo-electrons. The process basically, involves shining light of frequency $\upsilon$, on a metal surface to remove the electrons with a kinetic energy of $\frac{\text{1}}{\text{2}}\text{m}{v}^{\text{2}}$.

Answer to Problem 9.35E

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is $\text{4}{\text{.537×10}}^{\text{5}}{\text{ m sec}}^{-\text{1}}$.

Explanation of Solution

The given value of work function $\varphi$ is $\text{2}\text{.16 eV}$ and $\lambda$ is $\text{450nm}$.

The formula to calculate the velocity of an electron is given below.

$\begin{array}{rll}\frac{\text{hc}}{\lambda }& =& \varphi +\frac{1}{2}\text{m}{v}^2\cdot \cdot \cdot (2)\\ \end{array}$

Where,

• $\text{h}$ is the Planck’s Constant $\left(\text{6}{\text{.6×10}}^{-\text{34}}\text{ J sec}\right)$

• $\text{c}$ is the speed of light $\left(3.8\times {10}^8{\text{ m sec}}^{-\text{1}}\right)$

• $\lambda$ is the wavelength

• $\text{m}$ is the mass of an electron

• $v$ is the velocity of the emitted electron

• $\varphi$ is the work function

It is known that,

1eV $=\text{1}{\text{.6×10}}^{-\text{31}}\text{ J}$

Therefore,

$\text{2}\text{.16 }\text{eV×}\frac{\text{1}{\text{.6×10}}^{-\text{31}}\text{ J}}{\text{1 eV}}=3.456\times {10}^{-\text{19}}\text{ J}$

It is known that,

1nm ${=\text{10}}^{-\text{9}}\text{ m}$

Therefore,

$\frac{{\text{450 nm ×10}}^{-\text{9}}}{\text{1 nm}}={\text{450 ×10}}^{-\text{9}}\text{ nm}$

Rearrange equation (2) to calculate the value of velocity.

$v=\sqrt{\frac{\text{2}}{\text{m}}\left(\frac{\text{hc}}{\lambda }-\varphi \right)}$

Substitute the values of work function $\varphi$, Planck’s constant $\text{h}$, speed of light $\text{c}$, wavelength $\lambda$ and mass of electron $\text{m}$ in the above equation, we get-

$\begin{array}{rll}v& =& \sqrt{\frac{\text{2}\left(\frac{\text{6}{\text{.6×10}}^{-\text{34}}{\text{ J sec × 3×10}}^{\text{8}}{\text{ m sec}}^{-\text{1}}}{{\text{450×10}}^{-\text{9}}\text{ m}}-\text{3}{\text{.456×10}}^{-\text{19}}\text{ J}\right)}{\text{9}{\text{.1×10}}^{-\text{31}}\text{kg}}}\\ v& =& \sqrt{\frac{\text{2}\left(\text{0}{\text{.044×10}}^{-\text{17}}\text{ J}-\text{0}{\text{.03456×10}}^{-\text{17}}\text{ J}\right)}{\text{9}{\text{.1×10}}^{-\text{31}}\text{ kg}}}\\ v& =& \sqrt{\text{20}{\text{.586×10}}^{\text{10}}{\text{ m}}^{\text{2}}{\text{ sec}}^{-\text{2}}}\\ v& =& \text{4}{\text{.537 × 10}}^{\text{5}}{\text{ m sec}}^{-\text{1}}\end{array}$

Therefore, the value of speed is $4.537\times {10}^5{\text{ m sec}}^{-\text{1}}$.

Conclusion

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is $\text{4}{\text{.537×10}}^{\text{5}}{\text{ m sec}}^{-\text{1}}$

Interpretation Introduction

Interpretation:

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is to be stated.

Concept introduction:

The emission of electrons when the light is shone upon a material is called the photoelectric effect. The electrons emitted by such a method are also known as photo-electrons. The process basically involves shining light of frequency $\upsilon$, on a metal surface to remove the electrons with a kinetic energy of $\frac{\text{1}}{\text{2}}\text{m}{v}^{\text{2}}$.

Answer to Problem 9.35E

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is $\text{6}{\text{.86×10}}^{\text{5}}{\text{ m sec}}^{-\text{1}}$

Explanation of Solution

The given value of work function $\varphi$ is $\text{2}\text{.16 eV}$ and $\lambda$ is $\text{350nm}$.

The formula to calculate the velocity of an electron is given below.

$\begin{array}{rll}\frac{\text{hc}}{\lambda }& =& \varphi +\frac{1}{2}\text{m}{v}^2\cdot \cdot \cdot (3)\\ \end{array}$

Where,

• $\text{h}$ is the Planck’s Constant $\left(\text{6}{\text{.6×10}}^{-\text{34}}\text{ J sec}\right)$

• $\text{c}$ is the speed of light $\left(3.8\times {10}^8{\text{ m sec}}^{-\text{1}}\right)$

• $\lambda$ is the wavelength

• $\text{m}$ is the mass of an electron $v$ is the velocity of the emitted electron

• $\varphi$ is the work function

It is known that,

1eV $=\text{1}{\text{.6×10}}^{-\text{31}}\text{ J}$

Therefore,

$\text{2}\text{.16}\text{eV×}\frac{\text{1}{\text{.6×10}}^{-\text{31}}\text{ J}}{\text{1 eV}}=3.456\times {10}^{-\text{19}}\text{ J}$

It is known that,

1nm ${=\text{10}}^{-\text{9}}\text{m}$

Therefore,

$\frac{{\text{350 nm×10}}^{-\text{9}}}{\text{1 nm}}={\text{350×10}}^{-\text{9}}\text{ nm}$

Rearrange equation (3) to calculate the value of velocity.

$v=\sqrt{\frac{\text{2}}{\text{m}}\left(\frac{\text{hc}}{\lambda }-\varphi \right)}$

Substitute the values of work function $\varphi$, Planck’s constant $\text{h}$, speed of light $\text{c}$, wavelength $\lambda$ and mass of electron $\text{m}$ in the above equation.

$\begin{array}{rll}v& =& \sqrt{\frac{\text{2}\left(\frac{\text{6}{\text{.6×10}}^{-\text{34}}{\text{ J sec × 3×10}}^{\text{8}}{\text{ m sec}}^{-\text{1}}}{{\text{350×10}}^{-\text{9}}\text{ m}}-\text{3}{\text{.456×10}}^{-\text{19}}\text{ J}\right)}{\text{9}{\text{.1×10}}^{-\text{31}}\text{ kg}}}\\ v& =& \sqrt{\frac{\text{2}\left(\text{0}{\text{.05657×10}}^{-\text{17}}\text{ J}-\text{0}{\text{.03456×10}}^{-\text{17}}\text{ J}\right)}{\text{9}{\text{.1×10}}^{-\text{31}}\text{ kg}}}\\ v& =& \sqrt{\text{48}{\text{.37×10}}^{\text{10}}{\text{ m}}^{\text{2}}{\text{ sec}}^{-\text{2}}}\\ v& =& \text{6}{\text{.95×10}}^{\text{5}}{\text{ m sec}}^{-\text{1}}\simeq {\text{7×10}}^{\text{5}}{\text{ m sec}}^{-\text{1}}\end{array}$

Therefore, the value of speed is $\text{7}\times {10}^5{\text{msec}}^{-\text{1}}$.

Conclusion

The speed of an electron that is being emitted by rubidium, when light of different wavelength is shined on the metal in vacuum is ${\text{7×10}}^{\text{5}}{\text{ m sec}}^{-\text{1}}$