224k views
2 votes
Energy of photoelectrons emitted from a photosensitive surface is 1.56 ev. If it’s threshold wavelength is2500A find the wavelength of incident light

User Chris Poe
by
7.4k points

1 Answer

0 votes

Final answer:

To find the wavelength of the incident light, we can use the energy of the photoelectrons and the threshold wavelength. Using the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength, we can rearrange the equation to solve for the wavelength. By plugging in the values, we can calculate the wavelength of the incident light.

Step-by-step explanation:

The energy of photoelectrons emitted from a photosensitive surface is 1.56 eV and the threshold wavelength is 2500 Å. We can use the equation E = hc/λ to find the wavelength of the incident light.

First, we need to convert the energy from electron volts (eV) to joules (J). Since 1 eV is equal to 1.6 x 10^-19 J, the energy of the photoelectrons is 1.56 x 1.6 x 10^-19 J.

Next, we can rearrange the equation to solve for the wavelength:

λ = hc/E

Plugging in the values, λ = (6.626 x 10^-34 J·s × 3 x 10^8 m/s) / (1.56 x 1.6 x 10^-19 J)

Calculating this expression will give us the wavelength of the incident light.

User Captain Monk
by
7.6k points