Photons of energy 12eV are incident on a metal. It is found that current flows from the metal until a stopping potential of 8.0V is applied.

If the wavelength of the incident photons is doubled, what is the maximum kinetic energy of the ejected electrons?

What is KEmax in eV.

Please someone help me on this...thank you.

With 12 eV photons, photoelectrons are ejected with energies of up to 8 eV. The metal work function must be 4 volts.

Doubling the wavelength of the incident light decreases the photon energy to 6 eV. The work function remains 4 V. The maximum photoelectron energy is then 2 eV.

To find the maximum kinetic energy (KEmax) of the ejected electrons when the wavelength of the incident photons is doubled, we need to use the concept of the photoelectric effect and the equation:

KEmax = hf - Φ

where KEmax is the maximum kinetic energy of the ejected electrons, hf is the energy of the incident photons, and Φ is the work function of the metal.

Given that the energy of the incident photons (12eV) and the stopping potential (8.0V), we can calculate the work function of the metal (Φ) using the equation:

Φ = hf - KEmax

Since the energy of the incident photons (hf) remains constant when the wavelength is doubled, we can write:

12eV = hf

Therefore, hf = 12eV.

Now, substituting the values into the equation:

Φ = hf - KEmax

8.0V = 12eV - KEmax

Rearranging the equation to solve for KEmax:

KEmax = 12eV - 8.0V

KEmax = 4.0eV

So, the maximum kinetic energy (KEmax) of the ejected electrons is 4.0eV when the wavelength of the incident photons is doubled.

Please note that the given wavelength of the incident photons is not directly used in this calculation, as the energy is the essential parameter.

To answer this question, we need to understand the concept of the photoelectric effect and the equations associated with it. The photoelectric effect refers to the emission of electrons from a metal when it is exposed to light.

The energy of a photon is given by the equation:

E = hc/λ

where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the light.

In the photoelectric effect, electrons are ejected from the metal when the energy of the incident photons is greater than the work function of the metal (Φ). The work function represents the energy required to remove an electron from the metal surface.

The maximum kinetic energy of the ejected electrons (KEmax) is given by:

KEmax = E - Φ

where E is the energy of the incident photon and Φ is the work function.

In this case, we are given that the incident photons have an energy of 12 eV, and the stopping potential (Vstop) applied to the metal is 8.0 V. The stopping potential is the minimum potential required to stop all photoelectrons and prevent their further motion. Therefore, the stopping potential (Vstop) is equal to the maximum kinetic energy of the ejected electrons (KEmax) in electron volts (eV).

Using the equation KEmax = E - Φ, we can rearrange it to solve for Φ:

Φ = E - KEmax

Substituting the given values, we get:

Φ = 12 eV - 8.0 V

To solve this, we need to convert the electron volts (eV) to joules (J) since the work function (Φ) is typically expressed in joules. To convert eV to J, we can use the conversion factor:

1 eV = 1.602 x 10^-19 J

Converting 8.0 V to J, we multiply by the charge of an electron (e = 1.602 x 10^-19 C):

8.0 V * 1.602 x 10^-19 C = 1.282 x 10^-18 J

Now, we can substitute the values back into the equation:

Φ = 12 eV - 1.282 x 10^-18 J

To find the maximum kinetic energy of the ejected electrons (KEmax) when the wavelength is doubled, we need to determine the new energy of the incident photons. Doubling the wavelength represents halving the energy of the photons, since energy is inversely proportional to wavelength.

Therefore, the new energy (E2) of the incident photons is:

E2 = (12 eV)/2 = 6 eV

Substituting the new energy value into the rearranged equation:

Φ = 6 eV - 1.282 x 10^-18 J

To express the maximum kinetic energy in electron volts (eV), we can convert it back using the conversion factor:

1 J = 6.242 x 10^18 eV

Finally, the maximum kinetic energy of the ejected electrons (KEmax) in eV is:

KEmax = Φ = (6 eV - 1.282 x 10^-18 J) * (6.242 x 10^18 eV/J)

Calculating this expression will give you the value of KEmax in electron volts.