Question

Light with a wavelength of$\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}} \mathbf{m}$falls on the surface of a piece of chromium in an evacuated glass tube. If the work function of chromium is,$\mathbf{7}\mathbf{.}\mathbf{21}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{J}$ determine

(a) the maximum kinetic energy of the emitted photoelectrons and

(b) the speed of photoelectrons that have this maximum kinetic energy.

Short answer

1. The maximum kinetic energy of the emitted electrons is $7.4\times {10}^{-20}J$.
2. The speed of the ejected photoelectrons is $4.0\times {10}^{5}m{s}^{-1}$.

Step-by-step solution

Kinetic energy

It is the energy required by the body for its motion. It is defined as the energy desired for a body of a given mass to accelerate from interruption to its definite velocity.

Calculation

a.

The wavelength and energy of light are related by the formula as:

$E=\frac{hc}{\lambda }$

*h is Planck’s constant, i.e.,$6.625\times {10}^{-34}Js$.

*c is the velocity of light, i.e.,$3\times {10}^8 m{s}^{-1}$.

* E is the energy of the light.

The kinetic energy of the ejected electrons during photoelectric emission is given by the formula as:

$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\Phi$

*$\Phi$is the work function, i.e.,$7.21\times {10}^{-19}J$.

The kinetic energy of the ejected electrons are calculated by substituting in the above equation as:

$\begin{array}{rcl}\frac{1}{2}m{v}^2& =& \frac{hc}{\lambda }-\Phi \\ & =& \frac{6.625\times {10}^{-34}Js\times 3\times {10}^8 m{s}^{-1}}{2.50\times {10}^{-7}m}-7.21\times {10}^{-19}J\\ & =& 7.95\times {10}^{-19}J-7.21\times {10}^{-19}J\\ & =& 7.4\times {10}^{-20}J\end{array}$

Hence, the maximum kinetic energy of the emitted electrons is$7.4\times {10}^{-20}J$.

b.

The speed of the electrons can be calculated by using the kinetic energy formula as:

$E=\frac{1}{2}m{v}^2$

Upon rearranging and substituting the velocity is calculated as:

$v=\sqrt{\frac{2E}{m}}$

*m is the mass of photoelectron, i.e.,$9.1\times {10}^{-31}kg$.

*E is the energy of the emitted photoelectrons, i.e.,$7.4\times {10}^{-20}J$.

*v is the speed of the photoelectrons.

$\begin{array}{rcl}v& =& \sqrt{\frac{2E}{m}}\\ & =& \sqrt{\frac{2\times 7.4\times {10}^{-20}J}{9.1\times {10}^{-31}kg}}\\ & =& 39502m{s}^{-1}\\ & =& 4.0\times {10}^{5}m{s}^{-1}\\ & & \end{array}$

Therefore, the speed of the ejected photoelectrons is $4.0\times {10}^{5}m{s}^{-1}$.