Question
its initial direction.When light shines on potassium, the photoelectrons' maximum speed is $5.0 \times 10^{5} \mathrm{~m} / \mathrm{s}$. Find the light's wavelength.
Step 1
The work function is given by $5 = 2.30 \times 1.6 \times 10^{-19} \, \text{J}$, which simplifies to $3.68 \times 10^{-19} \, \text{J}$. Show more…
Show all steps
Your feedback will help us improve your experience
Narayan Hari and 86 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
When light shines on potassium, the photo electrons' maximum speed is $4.2 \times 10^{5} \mathrm{m} / \mathrm{s} .$ Find the light's wavelength.
The Atom
The Quantum Model of the Atom
The maximum energy of photoelectrons emitted from potassium is $2.1 \mathrm{eV}$ when illuminated by light of wavelength $3 \times 10^{-7} \mathrm{~m}$ and $0.5 \mathrm{eV}$ when the light wavelength is $5 \times 10^{-7} \mathrm{~m}$. Use these results to obtain values for Planck's constant and the minimum energy needed to free an electron from potassium.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD