Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find the wavelength of photons that eject \(0.100\,{\rm{eV}}\) electrons from potassium, given that the binding energy is \(2.24\,{\rm{eV}}\). Are these photons visible?

Short Answer

Expert verified

\(\lambda = 530\,{\rm{nm}}.\) Yes, this wavelength of the electrons ejected from potassium surface is in wavelength range.

Step by step solution

01

The longest-wavelength EM radiation can eject photon

The kinetic energy of the electron is given by

\(K{E_e} = hf - BE\) ...(1)

Here\(K{E_e}\)is the kinetic energy,\(h\)is the plank constant,\(f\)is the frequency of the EM radiation and\(BE\)is the binding energy.

Now we know that the wavelength of EM radiation is given by

\(\lambda = \frac{c}{f}\) ...(2)

Where\(c\)is the speed of light.

So equation becomes,

\(K{E_e} = \frac{{hc}}{\lambda } - BE\) ...(3)

02

Binding energy of the electron in the potassium

Here, the maximum kinetic energy of the ejected electron from potassium is

\(K{E_e} = 0.100\,{\rm{eV}}\)

Binding energy of the electron in the potassium is

\(BE = 2.24\,{\rm{eV}}\)

Since, we know

\(1.00\;\,{\rm{J}} = 6.242 \times {10^{18}}\,{\rm{eV}}\)

Therefore we get

\(\begin{aligned}{}K{E_e} &= 0.100\,{\rm{eV}} \times \frac{{1.00\;\,{\rm{J}}}}{{6.242 \times {{10}^{18}}\,{\rm{eV}}}}\\ &= 1.60 \times {10^{ - 20}}\,{\rm{J}}\end{aligned}\)

We also know that: Planks constant \(h = 4.13 \times {10^{ - 15}}\,{\rm{eV}}{\rm{.s}}\)

Speed of light \(c = 3 \times {10^8}\,{\rm{m/s}}\)

And

\(\begin{aligned}{}BE &= 2.24\,{\rm{eV}} \times \frac{{1.00\;\,{\rm{J}}}}{{6.242 \times {{10}^{18}}\,{\rm{eV}}}}\\ &= 35.9 \times {10^{ - 20}}\,{\rm{J}}\end{aligned}\)

Therefore, from equation, we get

\(\begin{aligned}{}\lambda &= \frac{{hc}}{{K{E_e} + BE}}\\ &= \frac{{6.626 \times {{10}^{ - 34}}\,{\rm{Js}} \times 3.00 \times {{10}^8}\,{\rm{m\;}}{{\rm{s}}^{{\rm{ - 1}}}}}}{{1.60 \times {{10}^{ - 20}}\,\;{\rm{J}} + 35.9 \times {{10}^{ - 20}}\,{\rm{J}}}}\\ &= 530 \times {10^{ - 9}}\,{\rm{m}}\\{\rm{ = 530\;}}\,{\rm{nm}}\end{aligned}\)

The approximate range of visible spectrum is ranging from\({\rm{400nm to 700nm}}\). Therefore, yes, the above calculated wavelength of the electrons ejected from potassium surface is in wavelength range.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free