   Chapter 29, Problem 15PE

Chapter
Section
Textbook Problem

Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 420-nm photons. Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device?

To determine

The time taken by the electrons to reach the detection device.

Explanation

Given:

Wavelength of radiation =λ=420 nm = 420×109 m

Speed of radiation =c=3×108 ms

Mass of electron =m=9.1×1031 kg

Distance traveled to reach the device =d=2.50 cm = 0.025 m

Time taken by electron to reach the device =t

Maximum velocity of ejected electron =vmax

Energy of the radiation =E=?

Binding Energy of electrons =Wo=2.71 eV

Plank's constant =h=4.14×1015 eV s

Kinetic energy of the electrons =KEmax

Formula Used:

Energy of radiation is given as

E=hcλ

Kinetic energy of the ejected electron is given as

KEmax=EWo

Kinetic energy is given as

KEmax=(0.5)mv2max

Time taken by the electron is given as

t=dvmax

Calculation:

Energy of radiation is given as

E=hcλ

Inserting the values

E=(4.14×1015 )(3×108)(420×109)E=2

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