A metal foil has a threshold frequency of 5.45 \times 10^{14} \text{ } \text{Hz}5.45×10 14 Hz. Which of the colors of visible l
ight have enough energy to eject electrons from this metal? Select all that are correct.Red
Orange
Yellow
Green
Blue
Indigo
Violet
Orange
Yellow
Green
Blue
Indigo
Violet
2 answers:
Answer:
Green, Blue, Indigo, Violet
Explanation:
According to Einstein's equation of photoelectric effect, the kinetic energy of emitted photoelectron is the difference between the energy of the incident photon and the work function of the metal. The work function of the metal referee to the minimum energy that must be supplied in order to eject an electron from a metal surface. The energy of the incident photon must exceed the work function of the metal.
When we look at the electromagnetic spectrum, only the selected colours have frequency above the threshold frequency as shown by the image attached below.
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Answer:
The answer is D. gamma rays
Explanation:
A radioactive atom can have three different types of emission:
alpha particles (α) = they have a mass of 4 amu and they have a very low penetrating power.
Beta particles (β) = they have 5x amu and they have an intermediate penetrating power
Gamma rays (γ) = they are not particles basically just energy so its mass is ≈ 0 and its penetrating power is higher
For this reason Gamma emissions (γ) has the smallest mass value.
Answer:
The new temperature will be 565.83 K.
Explanation:
Gay Lussac's law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that the pressure of the gas is directly proportional to its temperature. This means that if the temperature increases, the pressure will increase; or if the temperature decreases, the pressure will decrease.
In other words, Gay-Lussac's law states that when a gas undergoes a constant volume transformation, the ratio of the pressure exerted by the gas temperature remains constant:
When an ideal gas goes from a state 1 to a state 2, it is true:
In this case:
- P1= 180 kPa
- T1= 291 K
- P2= 350 kPa
- T2= ?
Replacing:
Solving:
T2= 565.83 K
<u><em>The new temperature will be 565.83 K.</em></u>