A metallic surface emits photo electrons when green light falls on it. But the metal does not emit when yellow light falls on it. Which of the following is correct when red light falls on the metallic surface?

If yellow does not exceed the threshold, red certainly wont.

To determine whether or not red light will cause the metallic surface to emit photoelectrons, we need to understand the concept of the photoelectric effect. The photoelectric effect is the phenomenon where electrons are emitted from a material when it absorbs photons of sufficient energy.

In this case, we know that the metallic surface emits photoelectrons when green light falls on it, but not when yellow light falls on it. This information suggests that the electrons in the metal require a minimum amount of energy to be excited and emitted. Green light photons have more energy compared to yellow light photons, enabling them to provide enough energy for the electrons to be emitted.

When red light falls on the metallic surface, we need to determine if the energy of red light photons is sufficient to cause electron emission. To find out, we can compare the energy of red photons to the energy of green and yellow photons.

The energy of a single photon can be calculated using the equation:

E = hc/λ

Where:
- E is the energy of a photon
- h is Planck's constant (h ≈ 6.626 × 10^-34 J·s)
- c is the speed of light (c ≈ 3.00 × 10^8 m/s)
- λ is the wavelength of the light

To compare the energies, we need to know the wavelengths of green, yellow, and red light.

Green light typically has a wavelength of around 520 nm (nanometers) which can be converted to meters: λgreen ≈ 520 × 10^-9 m.

Yellow light typically has a wavelength of around 580 nm (nanometers) which can be converted to meters: λyellow ≈ 580 × 10^-9 m.

Red light typically has a wavelength of around 650 nm (nanometers) which can be converted to meters: λred ≈ 650 × 10^-9 m.

Plugging the values into the equation, we can calculate the energies:

Egreen = hc/λgreen
Eyellow = hc/λyellow
Ered = hc/λred

Comparing the energies will allow us to determine if red light has sufficient energy to cause electron emission.