Question

What wavelength of light is necessary to produce photoelectron of speed $\mathbf{2}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^8\mathbf{}\frac{m}{s}$with a magnesium target?

Short answer

The wavelength of light necessary to produce photoelectron of speed $2\times {10}^8\frac{\text{m}}{\text{s}}$ with a magnesium target is $82.5\text{\hspace{0.17em}nm}$.

Step-by-step solution

Converting into joules.

The energy of the photon is, $\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ . The Kinetic energy of the photon from the zinc plate is given by: ${\mathrm{KE}}_{\max }=\mathrm{E}-\mathrm{\phi }$ Here $\mathrm{\phi }$ is the work function.

Now reforming the expression, we have,

$\mathbf{\lambda }\mathbf{=}\frac{\mathrm{hc}}{\frac{1}{2}{\mathbf{m\lambda }}^2\mathbf{+}\mathbf{f}}$

Because magnesium's work function potential is 3.7 eV, it must first be converted to Joules:

$(3.7\text{eV})\left(\frac{1.6\times {10}^{-14}\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}cV}}\right)=5.92\times {10}^{-11}\text{J}$

Determine the value of wavelength.

Substituting the value to calculate the value of wavelength,

$\begin{array}{c}\lambda =\frac{(6.63\times {10}^{-14}\text{\hspace{0.17em}Js})\left(3\times {10}^2\frac{\text{m}}{\text{s}}\right)}{\frac{1}{2}(9.1\times {10}^{-13}\text{\hspace{0.17em}kg}){\left(2\times {10}^6\frac{\text{m}}{\text{s}}\right)}^2+5.92\times {10}^{-17}\text{\hspace{0.17em}J}}\\ =8.25\times {10}^{-8}\text{\hspace{0.17em}m}\\ =82.5\text{nm}\end{array}$

Hence the value of wavelength is 82.5 nm.