Chapter 5, Problem 56A

Interpretation Introduction

Interpretation:

The minimum frequency of light is needed to eject photoelectron from the atom of platinum needs to be deermined.

Concept introduction:

 Light acts as a wave. so, it can be said as having a frequency and a wavelength.

Answer to Problem 56A

The minimum frequency of light is needed to eject photoelectron from the atom of $Pt$

$\left(v\right)=1.37\times {10}^{15}Hz$

Explanation of Solution

Given: Energy per photon $E_{photon}=9.08\times {10}^{-19}J$

Explanations Light acts as a wave. so, it can be said as having a frequency and a wavelength. The wavelength of light is the distance between corresponding points in two adjacent light cycles, and the frequency of light is the number of cycles of light that pass a given point per second.

As per the Planck-Einstein equation

$E_{photon}=hv$

Where $E$ is the energy of a photon, $h$ refers to Planck’s constant, and $v$ refers to frequency.

Now putting value

$h=6.626\times {10}^{-34 }Js$

$E_{photon}=9.08\times {10}^{-19}J$

Frequency can be calculated as follows:

$v=\frac{E}{h}=\frac{9.08\times {10}^{-19}\text{ J}}{6.626\times {10}^{-34}\text{ Js}}=1.37\times {10}^{15}\text{ Hz}$

Frequency $\left(v\right)=1.37\times {10}^{15}Hz$

Conclusion

The minimum frequency of light is needed to eject photoelectron from atom of $Pt$

$\left(v\right)=1.37\times {10}^{15}Hz$$$