When a photon of light hits the surface of a metal, a photoelectron may be emitted. Some of the photon's energy is required to free the electron from the surface. The remaining energy becomes the kinetic energy of the photoelectron. The minimum energy for the emission of an electron from the surface of a particular metal is 2.78E-19 Joules. What is the kinetic energy of electrons emitted from the surface of the metal when it is irradiated by electromagnetic radiation of wavelength 115 nanometers? Enter your answer in scientific notation in units of J (joules).

To find the kinetic energy of the photoelectron, we need to calculate the energy of the photon using its wavelength.

First, let's convert the wavelength from nanometers to meters:
115 nanometers = 115 × 10^(-9) meters

The energy of a photon can be calculated using the formula:
E = hc/λ

Where:
E is the energy of the photon,
h is Planck's constant (6.626 × 10^(-34) J·s),
c is the speed of light (299,792,458 m/s), and
λ is the wavelength of the electromagnetic radiation.

Let's substitute the values into the formula:

E = (6.626 × 10^(-34) J·s × 299,792,458 m/s) / (115 × 10^(-9) m)

Simplifying the expression:

E = (1.988 × 10^(-24) J·m) / (115 × 10^(-9) m)

E = 1.727 × 10^(-15) J

Now, to find the kinetic energy of the photoelectron, we subtract the minimum energy required (2.78 × 10^(-19) J) from the energy of the photon:

Kinetic Energy = Energy of the photon - Minimum energy required

Kinetic Energy = 1.727 × 10^(-15) J - 2.78 × 10^(-19) J

Simplifying the expression:

Kinetic Energy = 1.727 × 10^(-15) J

Therefore, the kinetic energy of electrons emitted from the surface of the metal when it is irradiated by electromagnetic radiation of wavelength 115 nanometers is 1.727 × 10^(-15) J.

To find the kinetic energy of electrons emitted from the surface of the metal, we need to calculate the energy of the incident photon and then subtract the minimum energy required to free the electron from the metal surface.

1. First, we need to find the energy of the incident photon using the formula:

Energy of a photon = (Planck's constant * Speed of light) / wavelength

The Planck's constant is h = 6.62607015 × 10^(-34) J·s
The speed of light is c = 2.998 × 10^8 m/s
The wavelength is given as 115 nanometers, which is 115 × 10^(-9) meters.

Plugging these values into the formula, we have:

Energy of a photon = (6.62607015 × 10^(-34) J·s * 2.998 × 10^8 m/s) / (115 × 10^(-9) m)

Calculating the numerator and denominator separately:

Energy of a photon = (19.8600107 × 10^(-26) J·m) / (115 × 10^(-9) m)

Simplifying:

Energy of a photon ≈ 0.17217352 × 10^(-17) J

2. Now, we need to subtract the minimum energy required to free the electron from the metal surface, which is given as 2.78 × 10^(-19) J.

Kinetic energy of the photoelectron = Energy of the photon - Minimum energy to free the electron

Kinetic energy of the photoelectron = 0.17217352 × 10^(-17) J - 2.78 × 10^(-19) J

Calculating the subtraction:

Kinetic energy of the photoelectron = (0.17217352 - 0.0278) × 10^(-17) J

Simplifying:

Kinetic energy of the photoelectron ≈ 0.14437352 × 10^(-17) J

Since the question asks for the answer in scientific notation, the final answer is:

Kinetic energy of the photoelectron ≈ 1.4437352 × 10^(-18) J